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Chapter 4 Quadratic Equations (Concepts)
This chapter provides a comprehensive and in-depth treatment of Quadratic Equations, a fundamental type of polynomial equation that significantly extends our algebraic capabilities beyond the linear equations studied previously. Defined formally, a quadratic equation in one variable (say, $x$) is any equation that can be written in the standard form: $$ \mathbf{ax^2 + bx + c = 0} $$ where $a$, $b$, and $c$ are real numbers, with the critical constraint that the leading coefficient $a$ cannot be zero ($a \neq 0$). If $a$ were zero, the $x^2$ term would vanish, reducing the equation to a linear one. The presence of the $x^2$ term is the defining characteristic. Our primary objective when faced with a quadratic equation is to find its roots or solutions – these are the specific values of the variable $x$ that make the equation a true statement.
Unlike linear equations which typically have one solution, quadratic equations can have up to two roots. This chapter meticulously explores several powerful methods for finding these roots:
- Factorization: This method leverages the concept that if the product of two factors is zero, then at least one of the factors must be zero. We attempt to express the quadratic polynomial $ax^2 + bx + c$ as a product of two linear factors, often using techniques like splitting the middle term (finding two numbers whose sum is $b$ and product is $ac$) or applying relevant algebraic identities (like $A^2 - B^2 = (A+B)(A-B)$). Once factorized into the form $(px+q)(rx+s) = 0$, we set each linear factor equal to zero ($px+q=0$ or $rx+s=0$) and solve for $x$ to find the roots.
- Completing the Square: This is a systematic algebraic technique used to manipulate the quadratic equation $ax^2 + bx + c = 0$ into the form $(x+k)^2 = m^2$. By rearranging terms and adding a specific constant to both sides (derived from the coefficient of $x$), we create a perfect square trinomial on one side. Taking the square root of both sides then leads to two linear equations ($x+k = m$ and $x+k = -m$) which yield the roots. This method is not only practical for solving but is also fundamentally important as it provides the basis for deriving the general quadratic formula.
- Quadratic Formula: Derived rigorously using the method of completing the square, this formula provides a direct and universal solution for the roots of any quadratic equation $ax^2 + bx + c = 0$. The roots are given explicitly by: $$ \mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} $$ The '$\pm$' symbol indicates that there are generally two potential roots, one obtained using the plus sign and the other using the minus sign. This formula is invaluable as it works regardless of whether the quadratic expression is easily factorable.
A crucial concept introduced for analyzing quadratic equations without necessarily finding the exact roots is the discriminant. Designated by $D$ (or sometimes $\Delta$), the discriminant is the quantity under the square root sign in the quadratic formula: $$ \mathbf{D = b^2 - 4ac} $$ The value of the discriminant reveals the nature of the roots of the quadratic equation:
- If $\mathbf{D > 0}$: The equation has two distinct real roots (corresponding to the $\pm$ in the formula yielding different values).
- If $\mathbf{D = 0}$: The equation has two equal real roots (often referred to as one repeated real root, as the $\pm$ term becomes zero).
- If $\mathbf{D < 0}$: The equation has no real roots. The square root of a negative number is not a real number, indicating the roots lie in the complex number system (though complex numbers are typically introduced later).
Learning to calculate $D$ and interpret its sign is a key skill for quickly understanding the type of solutions expected.
Similar to our work with linear equations, a significant portion of this chapter is dedicated to practical applications. We focus on translating various word problems – often involving scenarios related to geometric areas, number relations, ages, speed/distance/time calculations, or work rates – into the mathematical language of quadratic equations. Once formulated, these equations are solved using one of the learned methods (factorization, completing the square, or the quadratic formula). An essential final step involves verifying the solutions obtained and ensuring their validity within the original context of the problem (e.g., rejecting negative solutions if the variable represents a physical length or time). This chapter deepens algebraic problem-solving skills by equipping us to handle relationships that lead to quadratic forms.
Basic Terms Related to Quadratic Equations
In previous chapters, we explored linear equations, where the highest power of the variable is 1 (e.g., $3x + 5 = 0$). Now, we move to a new type of equation where the highest power of the variable is 2. These are known as quadratic equations, and they are fundamental in mathematics for modeling things like the path of a projectile, calculating areas, and in optimization problems.
Definition of a Quadratic Equation
An equation is called a quadratic equation in one variable if it can be written in the standard form:
$ax^2 + bx + c = 0$
where:
- $x$ is the variable.
- $a, b,$ and $c$ are real numbers called coefficients and the constant term.
- $a \neq 0$. This is the most important condition. If $a$ were 0, the $x^2$ term would vanish, and the equation would become a linear equation ($bx + c = 0$).
Components of a Quadratic Equation
In the standard form $ax^2 + bx + c = 0$:
- $ax^2$ is the quadratic term. The coefficient 'a' is the leading coefficient.
- $bx$ is the linear term.
- $c$ is the constant term or the absolute term.
Identifying Quadratic Equations
The key is to check if the equation, after simplification, can be written in the form $ax^2 + bx + c = 0$ with $a \neq 0$.
Examples of Quadratic Equations:
- $2x^2 - 5x + 3 = 0$ (Here $a=2, b=-5, c=3$)
- $x^2 - 9 = 0$ (Here $a=1, b=0, c=-9$)
- $4x^2 + x = 0$ (Here $a=4, b=1, c=0$)
- $(x+2)^2 = 0 \implies x^2 + 4x + 4 = 0$ (After expansion, it fits the standard form).
Examples of Equations that are NOT Quadratic:
- $5x - 10 = 0$ (This is a linear equation as the highest power of $x$ is 1).
- $x^3 - 2x^2 + 1 = 0$ (This is a cubic equation as the highest power of $x$ is 3).
- $x^2 + \frac{1}{x} = 5$ (This is not a polynomial equation because of the $\frac{1}{x}$ term).
Checking if an Equation is Quadratic
To determine if an equation is quadratic, you must first simplify it completely and arrange it into the standard form $ax^2 + bx + c = 0$.
Example 1. Check whether the following equation is a quadratic equation: $(x+1)^2 = 2(x-3)$.
Answer:
Solution
We start with the given equation: $(x+1)^2 = 2(x-3)$.
Expand both sides:
$x^2 + 2x + 1 = 2x - 6$
Bring all terms to the left-hand side (LHS) to set the equation to zero.
$x^2 + 2x + 1 - 2x + 6 = 0$
Combine like terms:
$x^2 + (2x - 2x) + (1 + 6) = 0$
$x^2 + 0x + 7 = 0$
This equation is in the form $ax^2 + bx + c = 0$, where $a=1, b=0, c=7$. Since $a=1 \neq 0$, the highest power of the variable is 2.
Answer: Yes, the given equation is a quadratic equation.
Example 2. Check whether the following equation is a quadratic equation: $x(x+1)+8 = (x+2)(x-2)$.
Answer:
Solution
We start with the given equation: $x(x+1)+8 = (x+2)(x-2)$.
Expand both sides. Use the identity $(a+b)(a-b) = a^2 - b^2$ on the right side.
$x^2 + x + 8 = x^2 - 2^2$
$x^2 + x + 8 = x^2 - 4$
Bring all terms to the LHS:
$x^2 - x^2 + x + 8 + 4 = 0$
Combine like terms:
$(x^2 - x^2) + x + (8 + 4) = 0$
$0x^2 + x + 12 = 0$
The simplified form is $x + 12 = 0$. Here, the coefficient of $x^2$ is $a=0$. Since the $x^2$ term has vanished, the highest power of the variable is 1.
Answer: No, the given equation is not a quadratic equation; it is a linear equation.
Example 3. Check whether the following is a quadratic equation: $x^2(x+2) = 8$.
Answer:
Solution
We start with the given equation: $x^2(x+2) = 8$.
Expand the left side:
$x^3 + 2x^2 = 8$
Bring all terms to the LHS to set the equation to zero.
$x^3 + 2x^2 - 8 = 0$
In this simplified form, the highest power of the variable $x$ is 3. For an equation to be quadratic, the highest power must be 2.
Answer: No, the given equation is not a quadratic equation; it is a cubic equation.
Roots (or Solutions) of a Quadratic Equation
A real number $\alpha$ is called a root or a solution of the quadratic equation $ax^2 + bx + c = 0$ if it satisfies the equation, meaning $a\alpha^2 + b\alpha + c = 0$.
In simpler terms, the roots are the specific values of $x$ that make the equation true. Finding the roots of a quadratic equation is the same as "solving" it.
The roots of a quadratic equation are the same as the zeros of the corresponding quadratic polynomial $p(x) = ax^2 + bx + c$.
Geometrical Meaning of Roots
Geometrically, the real roots of a quadratic equation are the x-coordinates of the points where the graph of the parabola $y = ax^2 + bx + c$ intersects the x-axis.
A quadratic equation can have:
- Two distinct real roots: The parabola intersects the x-axis at two different points.
- Two equal real roots (one distinct root): The parabola touches the x-axis at exactly one point (its vertex).
- No real roots: The parabola does not intersect the x-axis at all. (The roots in this case are complex numbers).
Relation Between Roots and Coefficients of a Quadratic Equation
For any quadratic equation, a powerful and direct relationship exists between its roots (the solutions) and its coefficients. These relationships allow us to find the sum and product of the roots without actually solving the equation, and conversely, to build a quadratic equation if we know its roots.
Formulas for Sum and Product of Roots
Consider the standard form of a quadratic equation:
$ax^2 + bx + c = 0$ (where $a \neq 0$)
Let the two roots of this equation be $\alpha$ (alpha) and $\beta$ (beta).
1. Sum of the Roots ($\alpha + \beta$)
The sum of the roots is given by the formula:
$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$
... (1)
2. Product of the Roots ($\alpha \cdot \beta$)
The product of the roots is given by the formula:
$\alpha\beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$
... (2)
Derivation of the Formulas
If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, then $(x - \alpha)$ and $(x - \beta)$ must be its factors.
Therefore, we can write the equation as:
$ax^2 + bx + c = k(x - \alpha)(x - \beta)$
where $k$ is some constant. By comparing the coefficient of the $x^2$ term on both sides, we can see that $k$ must be equal to $a$.
$ax^2 + bx + c = a(x - \alpha)(x - \beta)$
Now, expand the right-hand side:
$ax^2 + bx + c = a(x^2 - \beta x - \alpha x + \alpha\beta)$
$ax^2 + bx + c = a[x^2 - (\alpha + \beta)x + \alpha\beta]$
$ax^2 + bx + c = ax^2 - a(\alpha + \beta)x + a(\alpha\beta)$
For this equality to hold true for all values of $x$, the coefficients of corresponding powers of $x$ on both sides must be equal.
Comparing coefficients of $x$:
$b = -a(\alpha + \beta) \implies \alpha + \beta = -\frac{b}{a}$
Comparing constant terms:
$c = a(\alpha\beta) \implies \alpha\beta = \frac{c}{a}$
This completes the derivation of the formulas.
Forming a Quadratic Equation from its Roots
If you know the roots of a quadratic equation, you can construct the equation. If the roots are $\alpha$ and $\beta$, then we can write the equation as:
$x^2 - (\text{Sum of Roots})x + (\text{Product of Roots}) = 0$
Let $S = \alpha + \beta$ (Sum) and $P = \alpha\beta$ (Product). The formula is:
$x^2 - Sx + P = 0$
Note: Any non-zero multiple of this equation, such as $k(x^2 - Sx + P) = 0$, will have the same roots.
Worked Examples
Example 1. For the quadratic equation $3x^2 + 5x - 2 = 0$, find the sum and product of the roots without solving the equation.
Answer:
Solution
Compare the given equation with the standard form $ax^2 + bx + c = 0$.
Here, $a = 3, b = 5, c = -2$.
Sum of the roots ($\alpha + \beta$):
$\alpha + \beta = -\frac{b}{a} = -\frac{5}{3}$
Product of the roots ($\alpha\beta$):
$\alpha\beta = \frac{c}{a} = \frac{-2}{3}$
Answer: The sum of the roots is $-\frac{5}{3}$ and the product of the roots is $-\frac{2}{3}$.
Example 2. Find a quadratic equation whose roots are 2 and -5.
Answer:
Solution
Let the roots be $\alpha = 2$ and $\beta = -5$.
Step 1: Find the sum of the roots (S).
$S = \alpha + \beta = 2 + (-5) = -3$
Step 2: Find the product of the roots (P).
$P = \alpha\beta = (2)(-5) = -10$
Step 3: Use the formula $x^2 - Sx + P = 0$.
$x^2 - (-3)x + (-10) = 0$
$x^2 + 3x - 10 = 0$
Answer: The required quadratic equation is $x^2 + 3x - 10 = 0$.
Example 3. If the sum of the roots of the quadratic equation $kx^2 - 3x + 5 = 0$ is 1, find the value of $k$.
Answer:
Solution
For the equation $kx^2 - 3x + 5 = 0$, the coefficients are:
$a = k, b = -3, c = 5$.
The sum of the roots is given by the formula $-\frac{b}{a}$.
Sum of roots $= -\frac{-3}{k} = \frac{3}{k}$.
We are given that the sum of the roots is 1.
Therefore, we can set up the equation:
$\frac{3}{k} = 1$
Solving for $k$, we get:
$k = 3$
Answer: The value of $k$ is 3.
Example 4. If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 6x + 4 = 0$, find the value of $\alpha^2 + \beta^2$.
Answer:
Solution
For the equation $x^2 - 6x + 4 = 0$, we have $a=1, b=-6, c=4$.
First, find the sum and product of the roots:
Sum: $\alpha + \beta = -\frac{b}{a} = -\frac{-6}{1} = 6$.
Product: $\alpha\beta = \frac{c}{a} = \frac{4}{1} = 4$.
We need to find the value of $\alpha^2 + \beta^2$. We can express this in terms of $(\alpha + \beta)$ and $\alpha\beta$ using the algebraic identity:
$(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$
Rearranging this identity to solve for $\alpha^2 + \beta^2$ gives:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
Now substitute the values of the sum and product:
$\alpha^2 + \beta^2 = (6)^2 - 2(4)$
$\alpha^2 + \beta^2 = 36 - 8 = 28$
Answer: The value of $\alpha^2 + \beta^2$ is 28.
Example 5. Form a quadratic equation whose roots are the reciprocals of the roots of the equation $5x^2 + 2x - 3 = 0$.
Answer:
Solution
Let the roots of the given equation $5x^2 + 2x - 3 = 0$ be $\alpha$ and $\beta$.
From the given equation, we have:
Sum of roots: $\alpha + \beta = -\frac{2}{5}$
Product of roots: $\alpha\beta = -\frac{3}{5}$
The new equation has roots that are the reciprocals of $\alpha$ and $\beta$, so the new roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
To form the new quadratic equation, we need the sum and product of these new roots.
Sum of new roots (S'):
$S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta}$
Substitute the known values:
$S' = \frac{-2/5}{-3/5} = \frac{-2}{5} \times \frac{5}{-3} = \frac{2}{3}$
Product of new roots (P'):
$P' = \left(\frac{1}{\alpha}\right) \left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta}$
Substitute the known value:
$P' = \frac{1}{-3/5} = -\frac{5}{3}$
Now, form the new equation using the formula $x^2 - S'x + P' = 0$:
$x^2 - \left(\frac{2}{3}\right)x + \left(-\frac{5}{3}\right) = 0$
$x^2 - \frac{2}{3}x - \frac{5}{3} = 0$
To eliminate the fractions, multiply the entire equation by 3:
$3x^2 - 2x - 5 = 0$
Answer: The required quadratic equation is $3x^2 - 2x - 5 = 0$.
Solving a Quadratic Equation by Factorisation
One of the most direct algebraic methods for solving a quadratic equation is the factorisation method. This technique relies on breaking down the quadratic polynomial into a product of two simpler linear expressions (factors) and then applying the Zero Product Property.
The Zero Product Property
This is a fundamental rule in algebra that states: If the product of two or more numbers is zero, then at least one of those numbers must be zero.
For any real numbers A and B:
If $A \times B = 0$, then $A = 0$ or $B = 0$ (or both are 0).
When we factor a quadratic equation like $ax^2 + bx + c = 0$ into the form $(px + q)(rx + s) = 0$, we can treat $(px+q)$ as A and $(rx+s)$ as B. This allows us to find the roots by solving two simple linear equations: $px+q=0$ and $rx+s=0$.
Steps for Solving by Factorisation (Splitting the Middle Term)
- Standard Form: Ensure the equation is written as $ax^2 + bx + c = 0$.
- Find the Product and Sum: Calculate the product $ac$ (coefficient of $x^2$ times the constant term) and identify the sum $b$ (coefficient of $x$).
- Find Two Numbers: Find two numbers that multiply to give $ac$ and add to give $b$.
- Split the Middle Term: Rewrite the middle term, $bx$, as the sum of two terms using the numbers found in the previous step.
- Factor by Grouping: Group the first two terms and the last two terms, then factor out the greatest common factor from each group.
- Apply Zero Product Property: Set each linear factor equal to zero.
- Solve for x: Solve the two resulting linear equations to find the two roots of the quadratic equation.
Worked Examples
Example 1. Solve the quadratic equation $x^2 - 3x - 10 = 0$ by factorisation.
Answer:
Solution
Given the equation $x^2 - 3x - 10 = 0$.
Here, $a=1, b=-3, c=-10$.
We need two numbers that multiply to $ac = (1)(-10) = -10$ and add to $b = -3$.
The factors of -10 are (1, -10), (-1, 10), (2, -5), (-2, 5). The pair that adds to -3 is 2 and -5.
Split the middle term $-3x$ as $+2x - 5x$:
$x^2 + 2x - 5x - 10 = 0$
Factor by grouping:
$x(x + 2) - 5(x + 2) = 0$
Factor out the common binomial $(x+2)$:
$(x + 2)(x - 5) = 0$
Apply the Zero Product Property:
Either $x + 2 = 0$ or $x - 5 = 0$.
Solving each linear equation:
$x = -2$
$x = 5$
Answer: The roots are -2 and 5.
Example 2. Solve the quadratic equation $6x^2 - x - 2 = 0$ by factorisation.
Answer:
Solution
Given the equation $6x^2 - x - 2 = 0$.
Here, $a=6, b=-1, c=-2$.
We need two numbers that multiply to $ac = (6)(-2) = -12$ and add to $b = -1$.
The factors of -12 that add to -1 are -4 and 3.
Split the middle term $-x$ as $-4x + 3x$:
$6x^2 - 4x + 3x - 2 = 0$
Factor by grouping:
$2x(3x - 2) + 1(3x - 2) = 0$
Factor out the common binomial $(3x-2)$:
$(3x - 2)(2x + 1) = 0$
Apply the Zero Product Property:
Either $3x - 2 = 0$ or $2x + 1 = 0$.
Solving each linear equation:
$3x = 2 \implies x = \frac{2}{3}$
$2x = -1 \implies x = -\frac{1}{2}$
Answer: The roots are $\frac{2}{3}$ and $-\frac{1}{2}$.
Example 3. Solve the quadratic equation $3x^2 - 2\sqrt{6}x + 2 = 0$ by factorisation.
Answer:
Solution
Given the equation $3x^2 - 2\sqrt{6}x + 2 = 0$.
Here, $a=3, b=-2\sqrt{6}, c=2$.
We need two numbers that multiply to $ac = (3)(2) = 6$ and add to $b = -2\sqrt{6}$.
The two numbers are $-\sqrt{6}$ and $-\sqrt{6}$, since $(-\sqrt{6}) \times (-\sqrt{6}) = 6$ and $(-\sqrt{6}) + (-\sqrt{6}) = -2\sqrt{6}$.
Split the middle term:
$3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0$
To factor by grouping, we can rewrite $3$ as $(\sqrt{3})^2$, $2$ as $(\sqrt{2})^2$, and $\sqrt{6}$ as $\sqrt{3}\sqrt{2}$.
$(\sqrt{3}x)^2 - \sqrt{3}\sqrt{2}x - \sqrt{3}\sqrt{2}x + (\sqrt{2})^2 = 0$
$\sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0$
Factor out the common binomial:
$(\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0$
This can be written as $(\sqrt{3}x - \sqrt{2})^2 = 0$.
Apply the Zero Product Property (both factors are the same):
$\sqrt{3}x - \sqrt{2} = 0 \implies \sqrt{3}x = \sqrt{2} \implies x = \frac{\sqrt{2}}{\sqrt{3}}$ or $x = \frac{\sqrt{6}}{3}$.
Answer: The equation has two equal roots, $x = \frac{\sqrt{6}}{3}$.
Solving a Quadratic Equation by Completing the Square
The method of completing the square is a powerful technique that transforms any quadratic equation into a form that can be solved by taking the square root of both sides. This method always works, even when factorisation is difficult or impossible with integers, and it forms the basis for deriving the quadratic formula.
The Core Idea: Creating a Perfect Square Trinomial
The method relies on the algebraic identities for squaring binomials:
- $(x+k)^2 = x^2 + 2kx + k^2$
- $(x-k)^2 = x^2 - 2kx + k^2$
The goal is to manipulate the equation $ax^2 + bx + c = 0$ so that one side becomes a perfect square trinomial like $x^2 \pm 2kx + k^2$. Notice that the constant term, $k^2$, is always the square of half the coefficient of the $x$ term, $(2k/2)^2 = k^2$.
Steps for Completing the Square
- Standard Form: Start with the equation $ax^2 + bx + c = 0$.
- Move Constant: Move the constant term $c$ to the right side of the equation.
- Divide by 'a': If the leading coefficient $a$ is not 1, divide every term in the equation by $a$.
- Complete the Square: Take half of the new coefficient of the $x$ term, square it, and add this value to both sides of the equation.
- Factor: Factor the left side of the equation as a perfect square binomial.
- Take Square Root: Take the square root of both sides, remembering to include the plus-or-minus symbol ($\pm$) on the right side.
- Isolate x: Solve for $x$ to find the two roots.
Worked Examples
Example 1. Solve the quadratic equation $2x^2 - 5x + 3 = 0$ by completing the square.
Answer:
Solution
Given: $2x^2 - 5x + 3 = 0$.
Step 1 & 2: Move Constant.
$2x^2 - 5x = -3$
Step 3: Divide by 'a' (which is 2).
$x^2 - \frac{5}{2}x = -\frac{3}{2}$
Step 4: Complete the Square. The coefficient of $x$ is $-\frac{5}{2}$.
Half of it is $\frac{1}{2} \times (-\frac{5}{2}) = -\frac{5}{4}$.
Squaring this gives $(-\frac{5}{4})^2 = \frac{25}{16}$. Add this to both sides.
$x^2 - \frac{5}{2}x + \frac{25}{16} = -\frac{3}{2} + \frac{25}{16}$
Step 5: Factor. The left side is now a perfect square.
$\left(x - \frac{5}{4}\right)^2 = -\frac{24}{16} + \frac{25}{16} = \frac{1}{16}$
Step 6: Take Square Root.
$x - \frac{5}{4} = \pm \sqrt{\frac{1}{16}} = \pm \frac{1}{4}$
Step 7: Isolate x.
$x = \frac{5}{4} \pm \frac{1}{4}$
This gives two roots:
$x_1 = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$
$x_2 = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1$
Answer: The roots are $\frac{3}{2}$ and 1.
Example 2. Solve the equation $x^2 - 4x - 3 = 0$ by completing the square.
Answer:
Solution
Given: $x^2 - 4x - 3 = 0$.
Step 1 & 2: Move Constant.
$x^2 - 4x = 3$
Step 3: Divide by 'a'. Here $a=1$, so this step is not needed.
Step 4: Complete the Square. The coefficient of $x$ is -4.
Half of it is -2. Squaring this gives $(-2)^2 = 4$. Add 4 to both sides.
$x^2 - 4x + 4 = 3 + 4$
Step 5: Factor.
$(x - 2)^2 = 7$
Step 6: Take Square Root.
$x - 2 = \pm \sqrt{7}$
Step 7: Isolate x.
$x = 2 \pm \sqrt{7}$
Answer: The roots are $2 + \sqrt{7}$ and $2 - \sqrt{7}$.
Example 3. Solve the equation $x^2 + 2x + 5 = 0$ by completing the square.
Answer:
Solution
Given: $x^2 + 2x + 5 = 0$.
Step 1 & 2: Move Constant.
$x^2 + 2x = -5$
Step 3 & 4: Complete the Square. The coefficient of $x$ is 2.
Half of it is 1. Squaring this gives $(1)^2 = 1$. Add 1 to both sides.
$x^2 + 2x + 1 = -5 + 1$
Step 5: Factor.
$(x + 1)^2 = -4$
Step 6: Take Square Root.
$x + 1 = \pm \sqrt{-4}$
At this point, we encounter the square root of a negative number. The square of any real number is non-negative, so we cannot find a real number that is the square root of -4.
Answer: The equation has no real roots.
Solving a Quadratic Equation by Using the Formula
While factorisation and completing the square are valid methods for solving quadratic equations, the quadratic formula provides a direct and universal way to find the roots of any quadratic equation in standard form. The quadratic formula is derived from the completing the square method, making it applicable to all quadratic equations.
The Quadratic Formula
For a quadratic equation written in the standard form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$, the roots of the equation are given by the formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
This single formula yields both roots of the quadratic equation by considering the '+' and '-' signs before the square root. The expression under the square root, $b^2 - 4ac$, is called the discriminant of the quadratic equation, often denoted by the Greek letter Delta ($\Delta$) or simply $D$. The discriminant plays a key role in determining the nature of the roots (as we will see in the next section).
The two roots of the quadratic equation are explicitly given by:
The first root, $\alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$
The second root, $\beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$
Derivation of the Quadratic Formula
The quadratic formula is not magic; it is derived directly by applying the method of completing the square to the general quadratic equation $ax^2 + bx + c = 0$.
- Start with the general equation:
- Move the constant term to the right side:
- Divide by the leading coefficient 'a' to make the $x^2$ coefficient 1:
- Complete the square: Take half of the coefficient of the x-term ($\frac{b}{a}$), which is $\frac{b}{2a}$. Square it to get $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$. Add this term to both sides of the equation.
- Factor the left side and simplify the right side: The left side is now a perfect square. Find a common denominator for the right side.
- Take the square root of both sides: Remember to include the $\pm$ sign.
- Isolate x: Subtract $\frac{b}{2a}$ from both sides.
- Combine the fractions: Since they have a common denominator, combine them into a single fraction.
$ax^2 + bx + c = 0$
$ax^2 + bx = -c$
$x^2 + \frac{b}{a}x = -\frac{c}{a}$
$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$
$\left(x + \frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a^2}$
$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$
Simplify the square root on the right side:
$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$
$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
This final expression is the quadratic formula.
Worked Examples
Example 1. Solve the quadratic equation $x^2 + 4x - 5 = 0$ using the quadratic formula.
Answer:
Solution
Step 1: Standard Form. The equation is already in the form $ax^2 + bx + c = 0$.
Step 2: Identify Coefficients.
$a = 1, b = 4, c = -5$
Step 3: Calculate the Discriminant (D).
$D = b^2 - 4ac = (4)^2 - 4(1)(-5)$
$D = 16 - (-20) = 16 + 20 = 36$
Since $D=36 > 0$, there are two distinct real roots.
Step 4: Substitute into the Formula.
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{36}}{2(1)} = \frac{-4 \pm 6}{2}$
Step 5: Calculate the Roots.
Root 1 (using '+'): $x_1 = \frac{-4 + 6}{2} = \frac{2}{2} = 1$
Root 2 (using '-'): $x_2 = \frac{-4 - 6}{2} = \frac{-10}{2} = -5$
Answer: The roots are 1 and -5.
Example 2. Solve the quadratic equation $2x^2 + x - 4 = 0$ using the quadratic formula.
Answer:
Solution
Step 1: Standard Form. The equation is in standard form.
Step 2: Identify Coefficients.
$a = 2, b = 1, c = -4$
Step 3: Calculate the Discriminant (D).
$D = b^2 - 4ac = (1)^2 - 4(2)(-4)$
$D = 1 - (-32) = 1 + 32 = 33$
Since $D=33 > 0$, there are two distinct real roots.
Step 4: Substitute into the Formula.
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{33}}{2(2)} = \frac{-1 \pm \sqrt{33}}{4}$
Step 5: Calculate the Roots.
The number 33 is not a perfect square, so we leave the roots in this irrational form.
Root 1: $x_1 = \frac{-1 + \sqrt{33}}{4}$
Root 2: $x_2 = \frac{-1 - \sqrt{33}}{4}$
Answer: The roots are $\frac{-1 + \sqrt{33}}{4}$ and $\frac{-1 - \sqrt{33}}{4}$.
Example 3. Solve the quadratic equation $x^2 + 2x + 2 = 0$ using the quadratic formula.
Answer:
Solution
Step 1: Standard Form. The equation is in standard form.
Step 2: Identify Coefficients.
$a = 1, b = 2, c = 2$
Step 3: Calculate the Discriminant (D).
$D = b^2 - 4ac = (2)^2 - 4(1)(2)$
$D = 4 - 8 = -4$
Since the discriminant $D = -4$ is negative, the square root of the discriminant ($\sqrt{-4}$) is not a real number. Therefore, the equation has no real roots.
Answer: The equation has no real roots.
Example 4. Solve the equation $3x^2 - 6x = -1$ using the quadratic formula.
Answer:
Solution
Step 1: Standard Form. We must first rewrite the equation in standard form by moving all terms to one side.
$3x^2 - 6x + 1 = 0$
Step 2: Identify Coefficients.
$a = 3, b = -6, c = 1$
Step 3: Calculate the Discriminant (D).
$D = b^2 - 4ac = (-6)^2 - 4(3)(1)$
$D = 36 - 12 = 24$
Since $D=24 > 0$, there are two distinct real roots.
Step 4: Substitute into the Formula.
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{24}}{2(3)} = \frac{6 \pm \sqrt{24}}{6}$
Step 5: Calculate and Simplify the Roots.
We can simplify $\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4}\sqrt{6} = 2\sqrt{6}$.
$x = \frac{6 \pm 2\sqrt{6}}{6}$
Now, we can factor out a 2 from the numerator and simplify the fraction:
$x = \frac{2(3 \pm \sqrt{6})}{6} = \frac{\cancel{2}(3 \pm \sqrt{6})}{\cancel{6}_3} = \frac{3 \pm \sqrt{6}}{3}$
The two roots are:
Root 1: $x_1 = \frac{3 + \sqrt{6}}{3}$
Root 2: $x_2 = \frac{3 - \sqrt{6}}{3}$
Answer: The roots are $\frac{3 + \sqrt{6}}{3}$ and $\frac{3 - \sqrt{6}}{3}$.
Nature of Roots of a Quadratic Equation
The "nature" of the roots of a quadratic equation describes what kind of numbers the roots are: are they real and different? Are they real and equal? Or are they not real at all? We can determine this without actually solving the equation, using a special component of the quadratic formula called the discriminant.
The Discriminant
Recall the quadratic formula for the equation $ax^2 + bx + c = 0$:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
The expression inside the square root, $b^2 - 4ac$, is the discriminant. It is denoted by $D$ or $\Delta$.
$D = b^2 - 4ac$
The value of the discriminant "discriminates" or distinguishes between the three possible types of roots for a quadratic equation.
The Three Cases for the Nature of Roots
The entire character of the roots is determined by whether the discriminant $D$ is positive, zero, or negative.
Case 1: $D > 0$ (Discriminant is Positive)
If $D$ is a positive number, its square root ($\sqrt{D}$) is a real and non-zero value. This means the $\pm$ in the quadratic formula will produce two different results.
$x_1 = \frac{-b + \sqrt{D}}{2a}$ and $x_2 = \frac{-b - \sqrt{D}}{2a}$
Conclusion: The equation has two distinct real roots.
(Geometrically, the parabola intersects the x-axis at two different points.)
Case 2: $D = 0$ (Discriminant is Zero)
If $D$ is exactly zero, then $\sqrt{D} = \sqrt{0} = 0$. The $\pm$ part of the formula has no effect.
$x = \frac{-b \pm 0}{2a} = \frac{-b}{2a}$
Conclusion: The equation has two equal real roots. (Sometimes stated as one real root or repeated roots).
(Geometrically, the vertex of the parabola touches the x-axis at exactly one point.)
Case 3: $D < 0$ (Discriminant is Negative)
If $D$ is a negative number, its square root ($\sqrt{D}$) is not a real number. Since we cannot take the square root of a negative number in the real number system, the formula does not produce any real solutions.
Conclusion: The equation has no real roots. (The roots are complex or imaginary, which are studied in higher mathematics).
(Geometrically, the parabola is entirely above or entirely below the x-axis and never touches or crosses it.)
Summary Table
| Value of Discriminant ($D = b^2 - 4ac$) | Nature of Roots |
|---|---|
| $D > 0$ | Two distinct real roots |
| $D = 0$ | Two equal real roots |
| $D < 0$ | No real roots |
Worked Examples
Example 1. Find the nature of the roots of the quadratic equation $2x^2 - 4x + 3 = 0$.
Answer:
Solution
For the equation $2x^2 - 4x + 3 = 0$, we identify the coefficients:
$a = 2, b = -4, c = 3$.
Calculate the discriminant, $D = b^2 - 4ac$:
$D = (-4)^2 - 4(2)(3)$
$D = 16 - 24 = -8$
Since $D = -8$, which is less than 0 ($D < 0$), the equation falls into Case 3.
Answer: The quadratic equation has no real roots.
Example 2. Determine the nature of the roots of the equation $9x^2 - 6x + 1 = 0$.
Answer:
Solution
For the equation $9x^2 - 6x + 1 = 0$, we have:
$a = 9, b = -6, c = 1$.
Calculate the discriminant, $D = b^2 - 4ac$:
$D = (-6)^2 - 4(9)(1)$
$D = 36 - 36 = 0$
Since $D = 0$, the equation falls into Case 2.
Answer: The quadratic equation has two equal real roots.
Example 3. Find the value(s) of $k$ for which the quadratic equation $kx^2 - 2kx + 6 = 0$ has equal roots.
Answer:
Solution
For an equation to have equal roots, its discriminant must be zero ($D=0$).
For the equation $kx^2 - 2kx + 6 = 0$, the coefficients are:
$a = k, b = -2k, c = 6$.
Set up the discriminant and make it equal to zero:
$D = b^2 - 4ac = 0$
$(-2k)^2 - 4(k)(6) = 0$
Now, solve this equation for $k$:
$4k^2 - 24k = 0$
Factor out the common term $4k$:
$4k(k - 6) = 0$
Using the zero product property, we get two possible solutions for $k$:
$4k = 0 \implies k = 0$
$k - 6 = 0 \implies k = 6$
However, for the original equation to be a quadratic equation, the coefficient of $x^2$, which is $k$, cannot be zero ($k \neq 0$). Therefore, we must discard the solution $k=0$.
Answer: The value of $k$ for which the equation has equal roots is 6.
Application of Quadratic Equations to Solve Problems
Quadratic equations are essential tools for modeling and solving real-world problems. Situations involving area, speed and distance, projectile motion, and geometry often lead to quadratic equations. The key is to translate the words of the problem into a mathematical equation.
General Steps for Solving Word Problems
- Understand and Define: Read the problem carefully to understand the scenario. Identify the unknown quantity you need to find and assign it a variable (e.g., $x$). If there are other unknowns, express them in terms of $x$.
- Formulate the Equation: Use the information and relationships given in the problem to write an equation. This often involves using known formulas (like Area = length × width, or Distance = Speed × Time).
- Standard Form: Simplify the equation and arrange it into the standard quadratic form, $ax^2 + bx + c = 0$.
- Solve the Equation: Find the roots of the quadratic equation using the most appropriate method (factorisation, quadratic formula, etc.).
- Interpret the Roots: Check the solutions in the context of the original problem. A variable representing a physical quantity like length, time, or speed cannot be negative, so you may need to reject one of the roots.
- State the Final Answer: Clearly state the solution to the problem in a complete sentence with appropriate units.
Worked Examples
Example 1. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of an isosceles triangular park with its base as the breadth of the rectangle and an altitude of 12 m. Find the length and breadth of the rectangular park.
Answer:
Solution
Step 1 & 2: Define Variables.
Let the length of the rectangular park be $x$ metres.
Then, the breadth of the park is $(x - 3)$ metres.
(Note: For the breadth to be a positive number, we must have $x > 3$.)
Step 3: Formulate the Equation.
Area of rectangular park = Length × Breadth = $x(x - 3) = x^2 - 3x$.
For the triangular park:
Base = Breadth of rectangle = $(x - 3)$ m.
Altitude = 12 m.
Area of triangular park = $\frac{1}{2} \times \text{Base} \times \text{Altitude} \ $$ = \frac{1}{2} \times (x - 3) \times 12 \ $$ = 6(x - 3) = 6x - 18$.
The problem states: Area of rectangle = Area of triangle + 4.
$x^2 - 3x = (6x - 18) + 4$
Step 4: Write in Standard Form.
$x^2 - 3x = 6x - 14$
$x^2 - 3x - 6x + 14 = 0$
$x^2 - 9x + 14 = 0$
Step 5: Solve the Equation.
We can solve by factorisation. We need two numbers that multiply to 14 and add to -9. These numbers are -7 and -2.
$(x - 7)(x - 2) = 0$
The solutions are $x = 7$ or $x = 2$.
Step 6: Interpret the Roots.
We established that for the breadth to be positive, $x$ must be greater than 3.
- If $x = 2$, the breadth would be $2-3 = -1$, which is impossible. So, we reject $x=2$.
- If $x = 7$, the breadth would be $7-3 = 4$, which is a valid dimension.
Step 7: State the Final Answer.
The length is 7 metres and the breadth is 4 metres.
Answer: The length of the rectangular park is 7 metres and its breadth is 4 metres.
Example 2. The sum of the squares of two consecutive positive integers is 365. Find the integers.
Answer:
Solution
Step 1 & 2: Define Variables.
Let the first positive integer be $x$.
Since the integers are consecutive, the second positive integer is $(x + 1)$.
Step 3: Formulate the Equation.
The sum of their squares is 365.
$(x)^2 + (x+1)^2 = 365$
Step 4: Write in Standard Form.
$x^2 + (x^2 + 2x + 1) = 365$
$2x^2 + 2x + 1 = 365$
$2x^2 + 2x - 364 = 0$
Divide the entire equation by 2 to simplify it:
$x^2 + x - 182 = 0$
Step 5: Solve the Equation.
Using the quadratic formula, $a=1, b=1, c=-182$.
$D = b^2 - 4ac = 1^2 - 4(1)(-182) = 1 + 728 = 729$.
$x = \frac{-1 \pm \sqrt{729}}{2(1)} = \frac{-1 \pm 27}{2}$
The two roots are:
$x_1 = \frac{-1 + 27}{2} = \frac{26}{2} = 13$
$x_2 = \frac{-1 - 27}{2} = \frac{-28}{2} = -14$
Step 6: Interpret the Roots.
The problem asks for positive integers. Therefore, we must reject the negative root $x = -14$.
The valid solution is $x = 13$.
Step 7: State the Final Answer.
The first integer is $x = 13$.
The second integer is $x + 1 = 13 + 1 = 14$.
Answer: The two consecutive positive integers are 13 and 14.
Example 3. A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer:
Solution
Step 1 & 2: Define Variables.
Let the speed of the stream be $x$ km/h.
Speed of the boat in still water = 18 km/h.
Speed upstream (against the current) = $(18 - x)$ km/h.
Speed downstream (with the current) = $(18 + x)$ km/h.
Distance = 24 km.
Step 3: Formulate the Equation.
We use the formula Time = Distance / Speed.
Time taken to go upstream ($T_{up}$) = $\frac{24}{18-x}$ hours.
Time taken to return downstream ($T_{down}$) = $\frac{24}{18+x}$ hours.
The problem states that the upstream journey takes 1 hour more than the downstream journey.
$T_{up} = T_{down} + 1$
$\frac{24}{18-x} = \frac{24}{18+x} + 1$
Step 4: Write in Standard Form.
$\frac{24}{18-x} - \frac{24}{18+x} = 1$
Find a common denominator:
$\frac{24(18+x) - 24(18-x)}{(18-x)(18+x)} = 1$
$\frac{432 + 24x - 432 + 24x}{18^2 - x^2} = 1$
$\frac{48x}{324 - x^2} = 1$
$48x = 324 - x^2$
$x^2 + 48x - 324 = 0$
Step 5: Solve the Equation.
Using the quadratic formula, $a=1, b=48, c=-324$.
$D = 48^2 - 4(1)(-324) = 2304 + 1296 = 3600$.
$x = \frac{-48 \pm \sqrt{3600}}{2(1)} = \frac{-48 \pm 60}{2}$
The two roots are:
$x_1 = \frac{-48 + 60}{2} = \frac{12}{2} = 6$
$x_2 = \frac{-48 - 60}{2} = \frac{-108}{2} = -54$
Step 6: Interpret the Roots.
The speed of the stream cannot be negative. Therefore, we reject $x = -54$.
The valid solution is $x = 6$.
Step 7: State the Final Answer.
Answer: The speed of the stream is 6 km/h.