Question

The number of real roots of $\sqrt{x^2-x-6}+\sqrt{6x^2-5x-39}=\sqrt{x^2+4x-21}$ is?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\sqrt{x^2-x-6}+\sqrt{6x^2-5x-39}=\sqrt{x^{+}4x-21}$

squaring on both sides:

$⟹(x^2-x-6)+(6x^2-5x-39)+2\sqrt{(x^2-x-6)(6x^2-5x-39)}=x^2+4x-21$

$⟹2\sqrt{x^2-x-6}\sqrt{6x^2-5x-39}=-6x^2+10x+24$

Again, squaring on both sides,

$⟹4(x^2-x-6)(6x^2-5x-39)=(-6x^2+10x+24{)}^2$

$⟹24x^4-44x^3-280x^2+276x+936=36x^4-120x^3-188x^2+480x+576$

$⟹-12x^4+76x^3-96x^2-240x+360=0$

$⟹-4(x-3{)}^2(x-2\left)\right(3x+5)=0$

$\therefore x=3,2,\frac{-5}{3}$

But only $x=3$ satisfies the equation.

$\therefore$ The number of real roots is $1$.