Question

Find the roots of the following quadratic equation by using the quadratic formula
$(x^2+3x+2{)}^2-8(x^2+3x)-4=0$

• A. $0,1,3,4$,
• B. $0,1,-3,4$,
• C. $0,1,-3,-4$
• D. None of these

Answer 1

The correct option is C $0,1,-3,-4$

Let $y=x^2+3x+2$
$=>y-2=x^2+3x$
Therefore,
The quadratic equation is
$y^2-8(y-2)-4=0$
$=>y^2-8y+16-4=0$
$=>y^2-8y+12=0$
$=>y^2-6y-2y+12=0$
$=>y(y-6)-2(y-6)=0$
$=>(y-6)(y-2)=0$
$=>y=6$ and $y=2$
Thus putting the value of $y=6$ in equation (i) we get,
$6=x^2+3x+2$
$=>x^2+3x-4=0$
$=>x^2+4x-x-4=0$
$=>x(x+4)-1(x+4)=0$
$=>(x+4)(x-1)=0$
$=>x=-4$ and $x=1$
Putting the value of $y=2$ we get
$2=x^2+3x+2$
$=>x^2+3x=0$
$=>x(x+3)=0$
$=>x=0$ and $x=-3$