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  • The number of real roots of the equation  <img alt="\sqrt{x^2-4 x+3}+\sqrt{x^2-9}=\sqrt{4 x^2-14 x+6}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csqrt%7Bx%5E2-4%20x&plus;

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

 

Option: 1

3


Option: 2

1


Option: 3

2


Option: 4

0


Answers (1)

best_answer

\begin{aligned} & \sqrt{x^2-4 x+3}+\sqrt{x^2-9}=\sqrt{4 x^2-14 x+6} \\ & \sqrt{(x-3)(x-1)}+\sqrt{(x-3)(x+3)}=\sqrt{4 x^2-12 x-2 x+6} \\ & \Rightarrow \sqrt{x-3}(\sqrt{x-1}+\sqrt{x+3})=\sqrt{4 x(x-3)-2(x-3)} \\ & \Rightarrow \sqrt{x-3}(\sqrt{x-1}+\sqrt{x+3})=\sqrt{(4 x-2)(x-3)} \\ & \Rightarrow \sqrt{x-3}(\sqrt{x-1}+\sqrt{x+3}-\sqrt{2(2 x-1)})=0 \\ & \sqrt{x-3}=0 \text { or } \sqrt{x-1}+\sqrt{x+3}-\sqrt{2(2 x-1)}=0 \end{aligned}

\begin{aligned} & x=3 \text { or } \sqrt{x-1}+\sqrt{x+3}=\sqrt{2(2 x-1)} \\ & x-1+x+3+2 \sqrt{(x-1)(x+3)}=4 x-2 \\ & \Rightarrow 2 \sqrt{(x-1)(x+3)}=2 x-4 \\ & \Rightarrow(x-1)(x+3)=(x-2)^2 \\ & \Rightarrow x^2+2 x-3=x^2+4-4 x \\ & \Rightarrow 6 x=7 \end{aligned}

 x=\frac{7}{6}   (not possible) 

Number of real root = 1

 

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vinayak

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