Give (b−x)2−4(a−x)(c−x)=0⇒ b2+x2−2bx−4(ac−ax−cx+x2)=0⇒ b2+x2−2bx−4ac+4ax+4cx−4x2=0⇒ −3x2+x(4(a+c)−2b)+(b2−4ac)=0D= (4(a+c)−2b)2−4(−3)(b2−4ac)⇒ 4[4(a+c)2+b2−4b(a+c)+3b2−12ac]⇒ 4[4a2+4c2+8ac−4ba−4bc+4b2−12ac]⇒ 16[a2+b2+c2−ab−bc−ac]⇒ 8(2a2+2b2+2c2−2ab−2bc−2ca)⇒ 8((a−b)2+(b−c)2+(c−a)2)≥0Roots are real and Equal and distinct

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If $a, b, c$ are real then ${(b - x)}^{2} - 4 (a - x) (c - x) = 0$ will have always roots which are

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real and equal

real and distinct

imaginary

both 1 and 2

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e {(b - x)}^{2} - 4 (a - x) (c - x) = 0 \\ \Rightarrow b^{2} + x^{2} - 2 b x - 4 (a c - a x - c x + x^{2}) = 0 \\ \Rightarrow b^{2} + x^{2} - 2 b x - 4 a c + 4 a x + 4 c x - 4 x^{2} = 0 \\ \Rightarrow - 3 x^{2} + x (4 (a + c) - 2 b) + (b^{2} - 4 a c) = 0 \\ D = {(4 (a + c) - 2 b)}^{2} - 4 (- 3) (b^{2} - 4 a c) \\ \Rightarrow 4 [4 {(a + c)}^{2} + b^{2} - 4 b (a + c) + 3 b^{2} - 12 a c] \\ \Rightarrow 4 [4 a^{2} + 4 c^{2} + 8 a c - 4 b a - 4 b c + 4 b^{2} - 12 a c] \\ \Rightarrow 16 [a^{2} + b^{2} + c^{2} - a b - b c - a c] \\ \Rightarrow 8 (2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a) \\ \Rightarrow 8 ({(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}) \geq 0 \\ R o o t s a r e r e a l a n d E q u a l a n d d i s t i n c t \end{array}$