If a, b, c, d are real numbers such that a

Question

(B) Let $f(x) = (x-a)(x-c) + 2(x-b)(x-d)$.Since $f(x)$ is a quadratic polynomial with a positive leading coefficient $(3x^2)$,the roots are real and d

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Real and distinct

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(B) Let $f(x) = (x-a)(x-c) + 2(x-b)(x-d)$.Since $f(x)$ is a quadratic polynomial with a positive leading coefficient $(3x^2)$,the roots are real and distinct if the discriminant $D > 0$.Alternatively,consider the values of $f(x)$ at $x=a, b, c, d$:$f(a) = (a-a)(a-c) + 2(a-b)(a-d) = 2(a-b)(a-d)$. Since $a  0$.$f(b) = (b-a)(b-c) + 2(b-b)(b-d) = (b-a)(b-c)$. Since $b > a$ and $b  0$ and $(b-c) $f(c) = (c-a)(c-c) + 2(c-b)(c-d) = 2(c-b)(c-d)$. Since $c > b$ and $c  0$ and $(c-d) $f(d) = (d-a)(d-c) + 2(d-b)(d-d) = (d-a)(d-c)$. Since $d > a$ and $d > c$,$(d-a) > 0$ and $(d-c) > 0$,so $f(d) > 0$.Since $f(a) > 0$ and $f(b) Since $f(c)  0$,there exists a root in $(c, d)$.Thus,the equation has two real and distinct roots.

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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