Question

The number of real solution of
$\sqrt{x^2-4x+3}=\sqrt{x^2-9}=\sqrt{{4x}^2-14x+6}$

• A. one
• B. two
• C. three
• D. none of these

Answer 1

Answer: (C)

three

$\sqrt{x^2-4x+3}=\sqrt{x^2-9}=\sqrt{4x^2-14x+6}$

First consider $\sqrt{x^2-4x+3}=\sqrt{x^2-9}$

$\Rightarrow x^2-4x+3=x^2-9$

$\Rightarrow 4x=12$

$\Rightarrow x=3$

Now consider $\sqrt{x^2-9}=\sqrt{4x^2-14x+6}$

$\Rightarrow x^2-9=4x^2-14x+6$

$\Rightarrow 3x^2-14x+15=0$

$\Rightarrow 3x(x-3)-5(x-3)=0$

$\Rightarrow (3x-5)(x-3)=0$

$\Rightarrow x=3;x=\frac{5}{3}$

Now consider $\sqrt{x^2-4x+3}=\sqrt{4x^2-14x+6}$

$\Rightarrow x^2-4x+3=4x^2-14x+6$

$\Rightarrow 3x^2-10x+3=0$

$\Rightarrow (3x-1)(x-3)=0$

$\Rightarrow x=3;x=\frac{1}{3}$

Therefore, number of real roots are $3$