Question

If the equation $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$ has both roots equal, then
(a) b = ac
(b) $b=\frac{1}{2}(a+c)$
(c) b² = ac
(d) $b=\frac{2ac}{(a+c)}$

Answer 1

(c) b² = ac

$$\begin{gathered} \text{It is given that the roots of the equation}\left\{\right(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0\}\text{are equal}\text{.} \\ \therefore (b^2-4ac)=0 \\ \Rightarrow {\left\{2b(a+c)\right\}}^2-4(a^2+b^2)(b^2+c^2)=0 \\ \Rightarrow 4b^2(a^2+2ac+c^2)-4(a^2{b}^2+a^2{c}^2+b^4+b^2{c}^2)=0 \\ \Rightarrow 4a^2{b}^2+8a{b}^{2}c+4b^2{c}^2-4a^2{b}^2-4a^2{c}^2-4b^4-4b^2{c}^2=0 \\ \Rightarrow 8a{b}^{2}c-4a^2{c}^2-4b^4=0 \\ \Rightarrow 2a{b}^{2}c-a^2{c}^2-b^4=0 \\ \Rightarrow b^4-2ac{b}^2+a^2{c}^2=0 \\ \Rightarrow {(b^2-ac)}^2=0 \\ \Rightarrow b^2-ac=0 \\ \Rightarrow b^2=ac \end{gathered}$$