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
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