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 :
Solution
$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$
<br/><br/>$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$
<br/><br/>$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain
<br/><br/>or
<br/><br/>$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$
<br/><br/>$2 \sqrt{(x-1)(x+3)}=2 x-4$
<br/><br/>$x^{2}+2 x-3=x^{2}-4 x+4$
<br/><br/>$6 \mathrm{x}=7$
<br/><br/>$\mathrm{x}=7 / 6$ (rejected)
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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