The number of real roots of the equation x^2-4x+3+x^2-9=√4x^2-14x+6, is:

Question

The number of real roots of the equation $\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+6}$, is:

Accepted Answer

Write $x^2-4x+3=(x-1)(x-3)$ and $x^2-9=(x-3)(x+3)$.

Check $x=3$: both radicals on the left vanish, and the right side is $\sqrt{4(9)-14(3)+6}=\sqrt{36-42+6}=0$. So $x=3$ works.

For $x>3$, squaring $\sqrt{x-1}+\sqrt{x+3}=\sqrt{4x-2}$ (after factoring out $\sqrt{x-3}$) leads to $x=\dfrac76$, which is not in the domain $x\ge3$ --- rejected.

Hence there is exactly 1 real root.

Solutions · 1