Question

Find the nature of the roots of the following quadratic equation. If roots are real, find them
$2x^2-9x+9=0$

Answer 1

We have,

$2x^2-9x+9=0$

Comparing that,

$a{x}^2+bx+c=0$

Then,$a=2,b=-9,c=9$

We know that,

Nature of roots are

$D=b^2-4ac$

$={(-9)}^2-4\times 2\times 9$

$=81-72$

$=9>0$

Roots are real.

Find them,

$2x^2-9x+9=0$

$2x^2-(6+3)x+9=0$

$2x^2-6x-3x+9=0$

$2x(x-3)-3(x-3)=0$

$(x-3)(2x-3)=0$

If $x-3=0$ then, $x=3$

If $2x-3=0$ then, $x=\frac{3}{2}$

Hence, $x=3,\frac{3}{2}$ is the answer.