Question

In each of the following determine whether the given values are solution of the given equation or not
$\left(i\right)x^2-3x+2=0,x=2,x=-1\left(ii\right)x^2+x+1=0,x=0,x=1\left(iii\right)x^2-3\sqrt{3x}+6=0,x=\sqrt{3},x=-2\sqrt{3}\left(iv\right)x+\frac{1}{x}=\frac{13}{6},x=\frac{5}{6},x=\frac{5}{6},x=\frac{4}{3}\left(v\right)2x^2-x+9=x^2+4x+3,x=2,x=3\left(vi\right)x^2-\sqrt{2x}-4=0,x=-\sqrt{2},x=-2\sqrt{2}\left(vii\right)a^2{x}^2-3abx+2b^2=0,x=\frac{a}{b},x=\frac{b}{a}$

Answer 1

Put the values in the equation and if it is equal to zero, then it is a solution.

(i) $x=2$

$x^2-3x+2$
=$2^2-3\times 2+2$
=$4-6+2=0$
yes it is a solution

$x=-1$

=$-1^2-3\times -1+2$
$1+3+2=6$
No, it is not a solution

(ii) $x=0$

$0+0+1\ne 0$
not a solution

$x=1$
$1+1+1\ne 0$
not a solution

(iii)
$x=\sqrt{3}$
$3-9+6=0$
it is a solution

$x=-2\sqrt{3}$
$12+18+6\ne 0$
not a solution

(iv)No, none of them is the solution.

(v) By simplifying

$x^2-5x+6$
$x=2$
$4-10+6=0$
yes, it is a solution.

$x=3$
$9-15+6=0$
yes, it is a solution.

vi) $x=-\sqrt{2}$
$2+4-4\ne 0$
not a solution

$x=-2\sqrt{2}$
$8+4-4\ne 0$
not a solution

vii) $x=\frac{b}{a}$ is a solution
$x=\frac{a}{b}$ is not a solution