Question

Equation $(x+2)(x+3)(x+8)(x+12)=4x^2$ has

• A. four real & distinct roots
• B. two irrational roots
• C. two integer roots
• D. two imaginary roots

Answer 1

The correct option is A four real & distinct roots

Equation $:(x+2)(x+3)(x+8)(x+12)=4x^2$

Rearranging,
$(x+2)(x+12)(x+3)(x+8)=4x^2$
or, $(x^2+14x+24)(x^2+11x+24)=4x^2$

Put $y=x^2+24$ and we get,
$(y+14x)(y+11x)=4x^2$
or, $y^2+25xy+154x^2=4x^2$
or, $y^2+25xy+150x^2=0$
or, $(y+10x)(y+15x)=0$

Putting back the value of $y$, we get
$(x^2+24+10x)(x^2+24+15x)=0$
or, $(x+4)(x+6)(x^2+15x+24)=0$

$\therefore$ The roots are, $x=-4,x=-6$

Also, $x^2+15x+24=0$
Solving this we get $x=\frac{-15+\sqrt{129}}{2}$ or $x=\frac{-15-\sqrt{129}}{2}$
$\therefore$ There are four real and distinct roots