# Roots of an equation

The values of the variable satisfying the given equation are called roots of an equation.
Thus, x = $\alpha$, is a root of the equation f(x) = 0 , if f($\alpha$) = 0
Example : x = 1 is a roots of an equation $x^{3} - 6x^{2}$ + 11x - 6 = 0 , because
$1^{3} - 6 \times 1^{2} + 11 \times 1$ - 6
= 1 - 6 + 11 - 6
= - 5 + 5
= 0
$x^{3} - 6x^{2}$ + 11x - 6 = 0 when we put x =1 then x = 1 is a root of the equation.

Note : The roots of the equation depends on the degree of the polynomial.
If the degree of the polynomial is 1 then there is only 1 root of an equation.
If the degree of the polynomial is 2 then there are 2 roots of the equations.
If the degree of the polynomial is 3 then there are 3 roots of the equations.
If the degree of the polynomial is 4 then there are 4 roots of the equations and so on.

## Examples on roots of an equation

Check whether the given value of variable satisfies the equation or not. If satisfies then what will be your conclusion.
1) $x^{2}$ + 1 = 0 for x = -1
$x^{2}$ + 1 = $(-1)^{2}$ + 1 = 1 + 1 2 $\neq$ 0
So x = 1 is not a root of the given equation.

2) $9x^{2}$ + 4 = 0 for x = $\frac{2}{3}i$
= $9x^{2}$ + 4 = $9\times(\frac{2}{3}i)^{2}$ + 4
= $9\times \frac{4i^{2}}{9}$ + 4
= - 4 + 4 (since $i^{2}$ = -1)
= 0
So x = $\frac{2}{3}i$ is a root of an equation.

3) $x^{2}$ + 2x + 5 = 0 for x = -1 - 2i
$x^{2}$ + 2x + 5 = $(-1 - 2i)^{2}$ + 2(-1 - 2i) + 5
= 1 + 4i + 4$i^{2}$ - 2 - 4i + 5
= 1 - 4 - 2 + 5 (since $i^{2}$ = -1)
= 6 - 6
= 0
So, x = -1 - 2i is a root of the given equation.

4) $4x^{2}$ -12x + 25 = 0 for x = $\frac{3}{2}$ + 2i
$4x^{2}$ -12x + 25 = $4\times \left ( \frac{3}{2}+ 2i \right )^{2} - 12 \left ( \frac{3}{2}+ 2i \right )$ + 25

= $4 \left ( \frac{9}{4}+ 6i + 4i^{2}\right ) - \left ( 18+ 24i \right )$ + 25

= 9 + 24i - 16 - 18 - 24i + 25
= 34 - 34
= 0
So, x = $\frac{3}{2}$ + 2i is a root of the given equation.

11th grade math

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