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
From Relation between Roots and Coefficients of a Quadratic Equation to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Dec 01, 23 01:16 AM
Nov 30, 23 10:59 PM
Nov 30, 23 01:08 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.