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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
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Verified Answer
The correct answer is:
Real and distinct
$$
\begin{aligned}
& \text { Here, }(x-a)(x-c)+2(x-b)(x-d) \\
& =x^2-(a+c) x+a c+2\left[x^2-(b+d) x+b d\right] \\
& =x^2-(a+c) x+a c+2 x^2-2(b+d) x+2 b d \\
& =3 x^2-(a+2 b+c+2 d)+a c+2 b d \\
& \quad \quad D=b^2-4 a c \\
& =(a+2 b+c+2 d)^2-12(a c+2 b d) \\
& =[(a+2 d)+(2 b+c)]^2-12(a c+2 b d) \\
& =[(a+2 d)-(2 b+c)]^2+4(a+2 d)(2 b+c) . \quad-12(a c+2 b d) \\
& =[(a+2 d)-(2 b+c)]^2+4 a c+8 a b+16 b d \\
& =[(a+2 d)-(2 b+c)]^2-8 a c-8 b d+8 a b+8 c d \\
& =[(a+2 d)-(2 b+c)]^2+8(c-b)(d-a)
\end{aligned}
$$
As, $a < b < c < d,(c-b)(d-a)>0$
$\therefore$ Roots are real and distinct.
\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 b+c)]^2-12(a c+2 b d) \\
& =[(a+2 d)-(2 b+c)]^2+4(a+2 d)(2 b+c) . \quad-12(a c+2 b d) \\
& =[(a+2 d)-(2 b+c)]^2+4 a c+8 a b+16 b d \\
& =[(a+2 d)-(2 b+c)]^2-8 a c-8 b d+8 a b+8 c d \\
& =[(a+2 d)-(2 b+c)]^2+8(c-b)(d-a)
\end{aligned}
$$
As, $a < b < c < d,(c-b)(d-a)>0$
$\therefore$ Roots are real and distinct.
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