Relation between Roots and Coefficients of a Quadratic Equation

We will learn how to find the relation between roots and coefficients of a quadratic equation.

Let us take the quadratic equation of the general form ax^2 + bx + c = 0 where a (≠ 0) is the coefficient of x^2, b the coefficient of x and c, the constant term.

Let α and β be the roots of the equation ax^2 + bx + c = 0

Now we are going to find the relations of α and β with a, b and c.

Now ax^2 + bx + c = 0

Multiplication both sides by 4a (a ≠ 0) we get

4a^2x^2 + 4abx + 4ac = 0

(2ax)^2 + 2 * 2ax * b + b^2 – b^2 + 4ac = 0

(2ax + b)^2 = b^2 - 4ac

2ax + b = ± \(\sqrt{b^{2} - 4ac}\)

x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

Therefore, the roots of (i) are \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

Let α = \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) and β = \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)

Therefore,

α + β = \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) + \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)

α + β = \(\frac{-2b}{2a}\)

α + β = -\(\frac{b}{a}\)

α + β = -\(\frac{coefficient of x}{coefficient of x^{2}}\)

Again, αβ = \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) × \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)

αβ = \(\frac{(-b)^{2} - (\sqrt{b^{2} - 4ac)}^{2}}{4a^{2}}\)

αβ = \(\frac{b^{2} - (b^{2} - 4ac)}{4a^{2}}\)

αβ = \(\frac{4ac}{4a^{2}}\)

αβ = \(\frac{c}{a}\)

αβ = \(\frac{constant term}{coefficient of x^{2}}\)

Therefore, α + β = -\(\frac{coefficient of x}{coefficient of x^{2}}\) and αβ = \(\frac{constant term}{coefficient of x^{2}}\) represent the required relations between roots (i.e., α and β) and coefficients (i.e., a, b and c) of equation ax^2 + bx + c = 0.

 For example, if the roots of the equation 7x^2 - 4x - 8 = 0 be α and β, then

Sum of the roots = α + β = -\(\frac{coefficient of x}{coefficient of x^{2}}\) = -\(\frac{-4}{7}\) = \(\frac{4}{7}\).

and

the product of the roots = αβ = \(\frac{constant term}{coefficient of x^{2}}\) = \(\frac{-8}{7}\) = -\(\frac{8}{7}\).

Solved examples to find the relation between roots and coefficients of a quadratic equation:

Without solving the equation 5x^2 - 3x + 10 = 0, find the sum and the product of the roots.

Solution:

Let α and β be the roots of the given equation.

Then,

α + β = -\(\frac{-3}{5}\) = \(\frac{3}{5}\) and

αβ = \(\frac{10}{5}\) = 2

To find the conditions when roots are connected by given relations

Sometimes the relation between roots of a quadratic equation is given and we are asked to find the condition i.e., relation between the coefficients a, b and c of quadratic equation. This is easily done using the formula α + β = -\(\frac{b}{a}\) and αβ = \(\frac{c}{a}\). This will clear when you go through illustrative examples.

1. If α and β are the roots of the equation x^2 - 4x + 2 = 0, find the value of

(i) α^2 + β^2

(ii) α^2 - β^2

(iii) α^3 + β^3

(iv \(\frac{1}{α}\) + \(\frac{1}{ β }\)

Solution:

The given equation is x^2 - 4x + 2 = 0 ...................... (i)

According to the problem, α and β are the roots of the equation (i)

Therefore,

α + β = -\(\frac{b}{a}\) = -\(\frac{-4}{1}\) = 4

and αβ = \(\frac{c}{a}\) = \(\frac{2}{1}\) = 2

(i) Now α^2 + β^2 = (α + β)^2 - 2αβ = (4)^2 – 2 * 2 = 16 – 4 = 12.

(ii) α^2 - β^2 = (α + β)( α - β)

Now (α - β)^2 = (α + β)^2 - 4αβ = (4)^2 – 4 * 2 = 16 – 8 = 8

⇒ α - β = ± √8

⇒ α - β = ± 2√2

Therefore, α^2 - β^2 = (α + β)( α - β) = 4 * (± 2√2) = ± 8√2.

(iii) α^3 + β^3 = (α + β)^3 - 3αβ(α + β) = (4)^3 – 3 * 2 * 4 = 64 – 24 = 40.

(iv) \(\frac{1}{α}\) + \(\frac{1}{ β }\) = \(\frac{ α + β }{α β }\) = \(\frac{4}{2}\) = 2.

11 and 12 Grade Math 

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