### Quadratic Equation Question 8

#### Question 8 - 31 January - Shift 1

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

(1) 0

(2) 1

(3) 3

(4) 2

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Roots of equations

$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$

$=\sqrt{4(x-\frac{12}{4})(x-\frac{2}{4})}$

$\Rightarrow \sqrt{x-3}=0 \Rightarrow x=3$ which is in domain

or

$\sqrt{x-1}+\sqrt{x+3}=\sqrt{4 x-2}$

$2 \sqrt{(x-1)(x+3)}=2 x-4$

$x^{2}+2 x-3=x^{2}-4 x+4$

$6 x=7$

$x=7 / 6$

The number of real root is 1.