If \(\frac{b}{x-a}\) = \(\frac{x+a}{b}\) then the value of x in ter

1.	If \(\frac{b}{x-a}\) = \(\frac{x+a}{b}\) then the value of x in terms of a and b isA) \(\sqrt{a^2+b^2}\)B) ± \(\sqrt{a^2+b^2}\)C) - \(\sqrt{a^2+b^2}\)D) ± \(\sqrt{a^2-b^2}\)
Answer» Correct option is (B) \(\pm\sqrt{a^2+b^2}\) We have \(\frac{b}{x-a}=\frac{x+a}{b}\) \(\Rightarrow b^2=(x+a)(x-a)\) (By cross multiplication) \(\Rightarrow x^2-a^2=b^2\) \((\because(x+a)(x-a)=x^2-a^2)\) \(\Rightarrow x^2=a^2+b^2\) \(\Rightarrow x=\pm\sqrt{a^2+b^2}\) Correct option is B) ± \(\sqrt{a^2+b^2}\)

If \(\frac{b}{x-a}\) = \(\frac{x+a}{b}\) then the value of x in terms of a and b isA) \(\sqrt{a^2+b^2}\)B) ± \(\sqrt{a^2+b^2}\)C) - \(\sqrt{a^2+b^2}\)D) ± \(\sqrt{a^2-b^2}\)

Answer»

Correct option is (B) \(\pm\sqrt{a^2+b^2}\)

We have \(\frac{b}{x-a}=\frac{x+a}{b}\)

\(\Rightarrow b^2=(x+a)(x-a)\) (By cross multiplication)

\(\Rightarrow x^2-a^2=b^2\) \((\because(x+a)(x-a)=x^2-a^2)\)

\(\Rightarrow x^2=a^2+b^2\)

\(\Rightarrow x=\pm\sqrt{a^2+b^2}\)

Correct option is B) ± \(\sqrt{a^2+b^2}\)