# FORMATION OF QUADRATIC EQUATION WITH GIVEN ROOTS

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In my previous post, we solved quadratic equations using the factorization method.

But first,

## what is the root of a quadratic equation?

The root of a quadratic equation is simply the solution pair of a quadratic equation.

For example, $x^2+x-6=0$ has a solution pair of $(2$, $-3$)

Hence $2$ and $-3$ are said to be roots of $x^2+x-6=0$.

To derive the quadratic equation from roots,
You will use the formula:

$x^2-$(sum of roots)x+(product of roots)$=0$
👉This post is the extension of our series on quadratic equation👈

#### Example 1

If the roots of a quadratic equation are $(3$,$5)$, find the equation.

solution:

Remember that the formula is
$x^2-(sum of roots)x+(product of roots)=0$

The Sum of roots is $3+5=8$

Product of roots is $3\times5=15$

Hence,

$x^2-(8)x+(15)=0$

$x^2-8x+15=0$

#### Example 2

If the roots of a quadratic equation are

$(\frac{-1}{2}$, $2$), find the equation.

Solution:

Sum of roots is $\frac{-1}{2}+2=\frac{3}{2}$

Product of roots is $\frac{-1}{2}\times2=-1$

$x^2-(\frac{3}{2})x+(-1)=0$

$x^2-\frac{3x}{2}-1=0$

Taking the L.C.M

$\frac{2x^2-3x-2}{2}=0$

$2x^2-3x-2=0\times2$

$2x^2-3x-2=0$

#### Example 3

Find the quadratic equation whose roots is $\frac{2}{5}$ and $\frac{-1}{2}$

Solution:

Sum of roots is $\frac{2}{5}+\frac{-1}{2}=\frac{-1}{10}$

Product of roots is $\frac{2}{5}\times\frac{-1}{2}=\frac{-1}{5}$

Inserting the value in the formula

$x^2-(\frac{-1}{10})x+(\frac{-1}{5})=0$

$x^2+\frac{1x}{10}-\frac{1}{5}=0$

Taking the L.C.M

$\frac{10x^2+1x-2}{10}=0$

Cross multiply

$10x^2+1x-2=0\times10$

$10x^2+1x-2=0$

#### Example 4

Given that the roots of a quadratic equation are $18$ and $12$, find the equation.

Solution:

The Sum of roots is $18+12=30$

Product if roots is $18\times12=216$

$x^2-(sum of roots)x+(product of roots)=0$.

$x^2-(30)x+(216)=0$

$x^2-30x+216=0$

#### Example 5

Find the quadratic equation whose roots are $-6$ and $4$.

Solution:

The Sum of roots is $-6+4=-2$

Product of roots is $-6\times4=-24$

$x^2-(-2)x+(-24)=0$

$x^2+2x-24=0$

#### Example 6

Find the quadratic equation whose roots are $-3$ and $\frac{-2}{5}$.

Solution:

Sum of roots is $-3+(\frac{-2}{5})=\frac{-17}{5}$

Product of roots is $-3\times\frac{-2}{5}=\frac{6}{5}$

$x^2-(\frac{-17}{5})x+(\frac{6}{5})=0$

$x^2+\frac{17x}{5}+\frac{6}{5}=0$

Taking the L.C.M

$\frac{5x^2+17x+6}{5}=0$

Cross multiply

$5x^2+17x+6=0times s5$

$5x^2+17x+6=0$.

Related post

#### Example 7

If the roots of a quadratic equation are $\frac{-1}{3}$ and $\frac{2}{3}$, find the equation.

Solution:

Sums of roots is $\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}$

Product of roots is $\frac{-1}{3}\times\frac{2}{3}=\frac{-2}{9}$

$x^2-(sum of roots)x+(product of roots)=0$

$x^2-(\frac{1}{3})x+(\frac{-2}{9})=0$

$x^2-\frac{1x}{3}-(\frac{2}{9})=0$

Taking the L.C.M

$\frac{9x^2-3x-2}{9}=0$

Cross multiply

$9x^2-3x-2=0\times 9$

$9x^2-3x-2=0$.

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