AP EAMCET · Maths · Quadratic Equation
If \(a, b, c, d\) are real numbers such that \(a < b < c < d\), then the roots of the equation \((x-a)(x-c)+2(x-b)(x-d)=0\) are
- A Real and need not be distinct
- B Real and distinct
- C Non-real and distinct
- D Non-real and need not be distinct
Answer & Solution
Correct Answer
(B) Real and distinct
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Here, }(x-a)(x-c)+2(x-b)(x-d) \\ & =x^2-(a+c) x+a c+2\left[x^2-(b+d) x+b d\right] \\ & =x^2-(a+c) x+a c+2 x^2-2(b+d) x+2 b d \\ & =3 x^2-(a+2 b+c+2 d)+a c+2 b d \\ & \quad \quad D=b^2-4 a c \\ & =(a+2 b+c+2 d)^2-12(a c+2 b d) \\ & =[(a+2 d)+(2…
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