The roots of a quadratic equation are the values of the variable that satisfies the equation. They are also known as the ‘zeroes’ of the quadratic equation. For the equation ax2 + bx + c = 0 the two roots α and β are:
- ${\alpha =\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}}$
- β = ${\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$
For example, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2} because they satisfy the equation when their values are substituted.
When x = -3, (-3)2 + 5 × -(3) + 6 = 0
When x = -2, (-2)2 + 5 × -(2) + 6 = 0
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Without solving the equation 3x2 – 2x – 1 = 0, find whether x = 1 is a root or solution of this equation.
Solution:
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Substituting x = 1 in the given equation 3x2 – 2x – 1 = 0, we get
3(1)2 – 2 (1) – 1 = 0
=> 3 – 2 – 1 = 0
=> 3 – 3 = 0
Therefore, x = 1 is a solution of the given equation 3x2 – 2x – 1 = 0
How to Find the Roots of a Quadratic Equation
We can solve the quadratic equation to find its roots in different ways.
Using the Quadratic Formula
As we know α and β are the two roots of the quadratic equation, whose values can be determined using the quadratic formula:
(α, β) = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Let us verify if, indeed, the roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}
In the equation x2 + 5x + 6 = 0, a = 1, b = 5, c = 6
(α, β) = ${\dfrac{-5\pm \sqrt{\left( 5\right) ^{2}-4\times 1\times 6}}{-2\times 1}}$
= ${\dfrac{-5\pm \sqrt{25-24}}{2}}$
= ${\dfrac{-5+\sqrt{1}}{2}}$
= ${\dfrac{-5+1}{2}}$ and ${\dfrac{-5-1}{2}}$
= {-3, -2}
Thus, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}
By Factoring
Here we will factor the middle term ‘5x’ of the same equation x2 + 5x + 6 = 0
x2 + 5x + 6 = 0
=> x2 + 3x +2x + 6 = 0
=> x(x + 3) + 2(x + 3) = 0
=> (x + 3)(x + 2)= 0
=> x + 3 = 0 or x + 2 = 0
=> x = {-3, -2}
Thus, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}
By Completing the Square
Step 1: Isolating the x2 and x terms to one side of the equation
y = x2 + 5x + 6
=> y – 6 = x2 + 5x
Step 2: Add ${\left(\dfrac{b}{2}\right) ^{2}}$ to both sides of the equation
Here, b = ${\left( \dfrac{5}{2}\right) ^{2}}$ = 25/4
=> ${y-6+\dfrac{25}{4}=x^{2}+5x+\dfrac{25}{4}}$
Step 3: Factor the right side of the equation into a perfect square
=> ${y+\dfrac{1}{4}=\left( x+\dfrac{5}{2}\right) ^{2}}$
Step 4: Rewriting the equation in terms of y
=> ${y=\left( x+\dfrac{5}{2}\right) ^{2}-\dfrac{1}{4}}$
When the quadratic function f(x) = y = 0
${\left( x+\dfrac{5}{2}\right) ^{2}-\dfrac{1}{4}=0}$
=> ${\left( x+\dfrac{5}{2}\right) ^{2}=\dfrac{1}{4}}$
=> ${x+\dfrac{5}{2}=\pm \sqrt{\dfrac{1}{4}}}$
=> ${x=\left\{ +\dfrac{1}{2}-\dfrac{5}{2},-\dfrac{1}{2}-\dfrac{5}{2}\right\}}$
=> x= {-3, -2}
Thus, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}
By Graphing
Graphing the quadratic equation manually using a graphing calculator and then identifying the x-intercepts will give the roots of the quadratic equation.
The graph shows the two x-intercepts are (-2, 0) and (-3, 0). Thus the two roots of the quadratic equation are (-3, -2)
Nature of Roots of the Quadratic Equation
The nature of the roots of the quadratic equation depends on the value of the discriminant as follows:
- If b2 – 4ac > 0, the quadratic equation has 2 real solutions
- If b2 – 4ac < 0, the quadratic equation has no real solutions but two different complex or imaginary roots
- If b2 – 4ac = 0, the quadratic equation has one real root
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Find the value(s) of g if the quadratic equation 3y2 + gy + 2 = 0 has equal roots.
Solution:
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As we know, thequadratic equation ax2 + bx + c = 0 has equal roots if its discriminant b2 – 4ac = 0
Here, a = 3, b = g, and c = 2
Putting the value in the discriminant, we get
b2 – 4ac = 0
=> (g)2 – 4 × 3 × 2 = 0
=> g2– 24 = 0
=> g2 = 24
=>g = ${\pm 2\sqrt{6}}$
Thus, when the given quadratic equation has equal roots, g = ${2\sqrt{6}}$ or g = ${-2\sqrt{6}}$
Sum of the Roots
The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
α + β = ${-\dfrac{b}{a}}$
Derivation
α + β = ${\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}+\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$
= ${\dfrac{-b+\sqrt{b^{2}-4ac}-b-\sqrt{b^{2}-4ac}}{2a}}$
= ${-\dfrac{2b}{2a}}$ = ${-\dfrac{b}{a}}$
Product of the Roots
The product of the roots of a quadratic equation is equal to the constant term divided by the leading coefficient.
α.β = ${\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}×\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$
= ${\dfrac{b^{2}+b\sqrt{b^{2}-14ac}-b\sqrt{b^{2}-4ac}-\left( b^{2}-4ac\right) }{4a^{2}}}$
= ${\dfrac{b^{2}-b^{2}+4ac}{4a^{2}}}$
= ${\dfrac{4ac}{4a^{2}}}$ = ${\dfrac{c}{a}}$
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Find the sum and the product of the roots of the quadratic equation a2 – 5a – 14 = 0
Solution:
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Comparing the given quadratic equation with the standard form ax2 + bx + c = 0, here a = 1, b = -5, c = -14
As we know,
α + β = ${-\dfrac{b}{a}}$
= 1
Thus the sum of the roots of the quadratic equation a2 – 5a – 14 = 0 is 1
α.β = ${\dfrac{c}{a}}$
= -14
Thus the product of the roots of the quadratic equation a2 – 5a – 14 = 0 is -14
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Write the quadratic equation with the given roots {7, -2}
Solution:
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The quadratic equation can be obtained by multiplying the factors formed by the roots {7, -2}
The factors are (x – 3)(x + 2)
Thus, the quadratic equation is (x – 3)(x + 2) = 0
=> x2 + 2x – 3x – 6 = 0
=> x2 – x – 6 = 0
