The roots of ax2+2bx+c=0 and bx2−2acx+b=0 are simultaneously real, then

The roots of ax2+2bx+c=0 are real then Δ =4b2−4ac≥0 ⇒ b2≥ ac −−−(1)and the roots of bx2−2acx+b=0 are real thenΔ = (−2ac)2−4(b)(b)≥0⇒ 4ac−4b2≥0⇒ b2≤ ac−−−(2)from (1) & (2)b2=ac

The roots of ax2+2bx+c=0 and bx2−2acx+b=0 are simultaneously real, then

The roots of ax2+2bx+c=0 are real then Δ =4b2−4ac≥0 ⇒ b2≥ ac −−−(1)and the roots of bx2−2acx+b=0 are real thenΔ = (−2ac)2−4(b)(b)≥0⇒ 4ac−4b2≥0⇒ b2≤ ac−−−(2)from (1) & (2)b2=ac

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$T h e r o o t s o f a x^{2} + 2 b x + c = 0 a n d b x^{2} - 2 \sqrt{a c} x + b = 0 a r e s i m u l \tan e o u s l y r e a l, t h e n$

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$a = b = c = 0$

$a c = b^{2}$

$4 b^{2} = a c$

$a b = c^{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} T h e r o o t s o f a x^{2} + 2 b x + c = 0 a r e r e a l t h e n \\ Δ = 4 b^{2} - 4 a c \geq 0 \\ \Rightarrow b^{2} \geq a c - - - (1) \\ a n d t h e r o o t s o f b x^{2} - 2 \sqrt{a c} x + b = 0 a r e r e a l t h e n \\ Δ = {(- 2 \sqrt{a c})}^{2} - 4 (b) (b) \geq 0 \\ \Rightarrow 4 a c - 4 b^{2} \geq 0 \\ \Rightarrow b^{2} \leq a c - - - (2) \\ f r o m (1) & (2) \\ b^{2} = a c \end{array}$