The number of real roots of the equation (sqrt{x^2-4x+3}+sqrt{x^2-9}=sqrt{4x^2-14x+6}), is
(a) 3
(b) 2
(c) 0
(d) 1
\(\textbf{Solution}\)
\(\textbf{Step 1: Simplify the equation by taking out the common factor}\)
\(\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+6}\)
\(\Rightarrow\sqrt{(x-3)(x-1)}+\sqrt{(x-3)(x+3)}=\sqrt{2(2x^2-7x+3)}\)
\(\Rightarrow\sqrt{(x-3)(x-1)}+\sqrt{(x-3)(x+3)}=\sqrt{2(2x^2-6x-x+3)}\)
\(\Rightarrow\sqrt{(x-3)(x-1)}+\sqrt{(x-3)(x+3)}=\sqrt{2(2x(x-3)-1(x-3)}\)
\(\Rightarrow\sqrt{(x-3)(x-1)}+\sqrt{(x-3)(x+3)}=\sqrt{2((2x-1)(x-3)}\)
\(\Rightarrow\sqrt{(x-3)(x-1)}+\sqrt{(x-3)(x+3)}-\sqrt{2((2x-1)(x-3)}=0\)
\(\Rightarrow(\sqrt{(x-3)})(\sqrt{(x-1)}+\sqrt{(x+3)}-\sqrt{2((2x-1)})=0\)
\(\textbf{Step 2: Solve the equation}\)
\(\sqrt{x-3}=0\)Â or \(\sqrt{(x-1)}+\sqrt{(x+3)}-\sqrt{2((2x-1)}=0\)
\(\Rightarrow x=3\) or \(\sqrt{(x-1)}+\sqrt{(x+3)}-\sqrt{2((2x-1)}=0\)
\(\sqrt{(x-1)}+\sqrt{(x+3)}=\sqrt{2((2x-1)}\)
Squaring both sides we, get
\(x-1+x+3-2\sqrt{(x-1)(x+3)}=2(2x-1)\)
\(\Rightarrow 2x+2-2\sqrt{(x-1)(x+3)}=2(2x-1)\)
\(\Rightarrow x+1-\sqrt{(x-1)(x+3)}=(2x-1)\)
\(\Rightarrow -\sqrt{(x-1)(x+3)}=(x-2)\)
Squaring both sides, we get
\(\Rightarrow (x-1)(x+3)=(x-2)^2\)
\(\Rightarrow x^2+2x-3=x^2-4x+4\)
\(\Rightarrow 6x-3=4\)
\(\Rightarrow x=\frac{7}{6}\)
\(\textbf{Step 3: Check whether values of}\) \(\mathbf{x}\) \(\textbf{satisfying the domain of equation}\)
Values of \(x\) are \(3 \text{and } \frac{7}{6}\) but for \(x=\frac{7}{6}\) the expression \(\sqrt{4x^2-14x+6}\) is undefined.
Because, \(\sqrt{4x^2-14x+6}=\sqrt{2((2x-1)(x-3)}\)
\(=\sqrt{2((2(\frac{7}{6})-1)(\frac{7}{6}-3)}\)
\(=\sqrt{2((\frac{4}{3})(\frac{-11}{6})}\)
\(=\sqrt{-\frac{44}{9}}\)
[Which is undefined because the expression inside square root is negative, as we know square root function is defined for either zero or positive input only]
Thus, \(x=3\) is the only solution.
Therefore, the number of solution of the equation is only 1.
\(\textbf{Hence, the correct answer is option (d).}\)