# 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.

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