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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