Question

Find the roots of the following equation:

($x^2+x$)($x^2+x-7$)+10=0

0x2+x)(x2+x−7)+10=0(x2+x)(x2+x−7)+10=, then

• A. $x=\frac{-1\pm \sqrt{21}}{2},1,-2$
• B. $x=\frac{-1\pm \sqrt{23}}{2},1,-2$
• C. $x=\frac{-1\pm \sqrt{21}}{2},-1,-2$
• D. $x=\frac{-1\pm \sqrt{21}}{2},1,2$

Answer 1

The correct option is A $x=\frac{-1\pm \sqrt{21}}{2},1,-2$

$(x^2+x)(x^2+x-7)+10=0$
Let $x^2+x=y$............(1)
Then , $y(y-7)+10=0$
$y^2-7y+10=0$
$(y-5)(y-2)=0$
$\therefore y=2,5$
Now substituting value of $y$ in equation (!), we get

$x^2+x-2=0$
$x=\frac{-1\pm \sqrt{1+8}}{2}=\frac{-1\pm 3}{2}$
$x=1,-2$
and, $x^2+x-5=0$
$x=\frac{-1\pm \sqrt{1+21}}{2}=\frac{-1\pm \sqrt{21}}{2}$
Thus, $x=\frac{-1\pm \sqrt{21}}{2},-1,-2$