Formation of the Quadratic Equation whose Roots are Given

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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