JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 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 :
0
1
3
2
Explanation
$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$
$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$
$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain
or
$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$
$2 \sqrt{(x-1)(x+3)}=2 x-4$
$x^{2}+2 x-3=x^{2}-4 x+4$
$6 \mathrm{x}=7$
$\mathrm{x}=7 / 6$ (rejected)
$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$
$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain
or
$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$
$2 \sqrt{(x-1)(x+3)}=2 x-4$
$x^{2}+2 x-3=x^{2}-4 x+4$
$6 \mathrm{x}=7$
$\mathrm{x}=7 / 6$ (rejected)
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