Question

Solve the following equation:
$2x^2+3x-5\sqrt{2x^2+3x+9}+3=0$

Answer 1

$2x^2+3x-5\sqrt{2x^2+3x+9}+3=0$

Let $2x^2+3x=t$

$\Rightarrow t-5\sqrt{t+9}+3=0$

$\Rightarrow t+3=5\sqrt{t+9}$

Lets do squaring on both sides,

$\Rightarrow {(t+3)}^2=25(t+9)$

$\Rightarrow t^2+9+6t=25t+225$

$\Rightarrow t^2+9+6t-25t-225=0$

$\Rightarrow t^2-19t-216=0$

Lets compare with $a{t}^2+bt+c=0$,

$a=1,b=-19$ and $c=-216$

$\Rightarrow$ The roots of $a{x}^2+bx+c=0$ are,

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow t=\frac{19\pm \sqrt{361-(4\times 1\times (-216\left)\right)}}{2}$

$\Rightarrow t=\frac{19\pm \sqrt{361+864}}{2}$

$\Rightarrow t=\frac{19\pm \sqrt{1225}}{2}$

$\Rightarrow t=\frac{19\pm 15}{2}$

Case1:

Lets take, $t=\frac{19+15}{2}$

$\Rightarrow t=\frac{19+15}{2}=\frac{34}{2}=17$

$\Rightarrow 2x^2+3x=17$

$\Rightarrow 2x^2+3x-17=0$

$\Rightarrow x=\frac{-3\pm \sqrt{9+136}}{4}$

$\Rightarrow x=\frac{-3\pm \sqrt{144}}{4}$

$\Rightarrow x=\frac{-3\pm 12}{4}$

$\Rightarrow x=\frac{-3+12}{4},x=\frac{-3-12}{4}$

$\Rightarrow x=\frac{9}{4},x=\frac{-15}{4}$

Case2:

Lets take $t=\frac{19-15}{2}$

$t=\frac{19-15}{2}=\frac{4}{2}=2$

$2x^2+3x=2$

$\Rightarrow 2x^2+3x-2=0$

$\Rightarrow 2x(x+2)-(x+2)=0$

$\Rightarrow (x+2)(2x-1)=0$

$\Rightarrow x+2=0,(2x-1)=0$

$\Rightarrow x=-2,x=\frac{1}{2}$

Therefore, $x=\frac{9}{4},\frac{-15}{4},-2,\frac{1}{2}$