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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