We will learn the formation of the quadratic equation whose roots are given.
To form a quadratic equation, let α and β be the two roots.
Let us assume that the required equation be ax\(^{2}\) + bx + c = 0 (a ≠ 0).
According to the problem, roots of this equation are α and β.
Therefore,
α + β = - \(\frac{b}{a}\) and αβ = \(\frac{c}{a}\).
Now, ax\(^{2}\) + bx + c = 0
⇒ x\(^{2}\) + \(\frac{b}{a}\)x + \(\frac{c}{a}\) = 0 (Since, a ≠ 0)
⇒ x\(^{2}\) - (α + β)x + αβ = 0, [Since, α + β = -\(\frac{b}{a}\)
and αβ = \(\frac{c}{a}\)]
⇒ x\(^{2}\) - (sum of the roots)x + product of the roots = 0
⇒ x\(^{2}\) - Sx + P = 0, where S = sum of the roots and P = product of the roots ............... (i)
Formula (i) is used for the formation of a quadratic equation when its roots are given.
For example suppose we are to form the quadratic equation whose roots are 5 and (-2). By formula (i) we get the required equation as
x\(^{2}\) - [5 + (-2)]x + 5 ∙ (-2) = 0
⇒ x\(^{2}\) - [3]x + (-10) = 0
⇒ x\(^{2}\) - 3x - 10 = 0
Solved examples to form the quadratic equation whose roots are given:
1. Form an equation whose roots are 2, and - \(\frac{1}{2}\).
Solution:
The given roots are 2 and -\(\frac{1}{2}\).
Therefore, sum of the roots, S = 2 + (-\(\frac{1}{2}\)) = \(\frac{3}{2}\)
And tghe product of the given roots, P = 2 ∙ -\(\frac{1}{2}\) = - 1.
Therefore, the required equation is x\(^{2}\) – Sx + p
i.e., x\(^{2}\) - (sum of the roots)x + product of the roots = 0
i.e., x\(^{2}\) - \(\frac{3}{2}\)x – 1 = 0
i.e, 2x\(^{2}\) - 3x - 2 = 0
2. Find the quadratic equation with rational coefficients which has \(\frac{1}{3 + 2√2}\) as a root.
Solution:
According to the problem, coefficients of the required quadratic equation are rational and its one root is \(\frac{1}{3 + 2√2}\) = \(\frac{1}{3 + 2√2}\) ∙ \(\frac{3 - 2√2}{3 - 2√2}\) = \(\frac{3 - 2√2}{9 - 8}\) = 3 - 2√2.
We know in a quadratic with rational coefficients irrational roots occur in conjugate pairs).
Since equation has rational coefficients, the other root is 3 + 2√2.
Now, the sum of the roots of the given equation S = (3 - 2√2) + (3 + 2√2) = 6
Product of the roots, P = (3 - 2√2)(3 + 2√2) = 3\(^{2}\) - (2√2)\(^{2}\) = 9 - 8 = 1
Hence, the required equation is x\(^{2}\) - Sx + P = 0 i.e., x\(^{2}\) - 6x + 1 = 0.
2. Find the quadratic equation with real coefficients which has -2 + i as a root (i = √-1).
Solution:
According to the problem, coefficients of the required quadratic equation are real and its one root is -2 + i.
We know in a quadratic with real coefficients imaginary roots occur in conjugate pairs).
Since equation has rational coefficients, the other root is -2 - i
Now, the sum of the roots of the given equation S = (-2 + i) + (-2 - i) = -4
Product of the roots, P = (-2 + i)(-2 - i) = (-2)\(^{2}\) - i\(^{2}\) = 4 - (-1) = 4 + 1 = 5
Hence, the required equation is x\(^{2}\) - Sx + P = 0 i.e., x\(^{2}\) - 4x + 5 = 0.
11 and 12 Grade Math
From Formation of the Quadratic Equation whose Roots are Given 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.