The equation x + √(x - 2) = 4 has A) two real rootsB) two imaginar

1.	The equation x + √(x - 2) = 4 has A) two real rootsB) two imaginary roots C) one real root D) one real and one imaginary root
Answer» Correct option is (C) one real root Given equation is \(x+\sqrt{x-2}=4\) _______________(1) \(\Rightarrow\) \(\sqrt{x-2}=4-x\) \(\Rightarrow x-2=(4-x)^2\) \(\Rightarrow x-2=x^2-8x+16\) \(\Rightarrow x^2-9x+18=0\) \(\Rightarrow x^2-6x-3x+18=0\) \(\Rightarrow x(x-6)-3(x-6)=0\) \(\Rightarrow(x-6)(x-3)=0\) \(\Rightarrow x-6=0\) or \(x-3=0\) \(\Rightarrow x=6\) or \(x=3\) Put x = 3 in equation (1), we have \(3+\sqrt{3-2}=4\) \(\Rightarrow\) 3+1 = 4 \(\Rightarrow\) 4 = 4 (Satisfies) Hence, x = 3 is a root of given equation. Put x = 6 in equation (1), we have \(6+\sqrt{6-2}=4\) \(\Rightarrow\) 6+2 = 4 \(\Rightarrow\) 8 = 4 (Not satisfies) \(\therefore\) x = 6 is not a root of given equation. Hence, x = 3 is only real root of given equation. Thus, given equation has only one real root. Correct option is C) one real root

The equation x + √(x - 2) = 4 has A) two real rootsB) two imaginary roots C) one real root D) one real and one imaginary root

Answer»

Correct option is (C) one real root

Given equation is \(x+\sqrt{x-2}=4\) _______________(1)

\(\Rightarrow\) \(\sqrt{x-2}=4-x\)

\(\Rightarrow x-2=(4-x)^2\)

\(\Rightarrow x-2=x^2-8x+16\)

\(\Rightarrow x^2-9x+18=0\)

\(\Rightarrow x^2-6x-3x+18=0\)

\(\Rightarrow x(x-6)-3(x-6)=0\)

\(\Rightarrow(x-6)(x-3)=0\)

\(\Rightarrow x-6=0\) or \(x-3=0\)

\(\Rightarrow x=6\) or \(x=3\)

Put x = 3 in equation (1), we have

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

\(\Rightarrow\) 3+1 = 4

\(\Rightarrow\) 4 = 4 (Satisfies)

Hence, x = 3 is a root of given equation.

Put x = 6 in equation (1), we have

\(6+\sqrt{6-2}=4\)

\(\Rightarrow\) 6+2 = 4

\(\Rightarrow\) 8 = 4 (Not satisfies)

\(\therefore\) x = 6 is not a root of given equation.

Hence, x = 3 is only real root of given equation.

Thus, given equation has only one real root.

Correct option is C) one real root