If a

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(A) Let $f(x) = (x - a)(x - c) + 2(x - b)(x - d)$.We evaluate the function at the given points $a, b, c, d$:$f(a) = (a - a)(a - c) + 2(a - b)(a - d) =

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Real and distinct

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(A) Let $f(x) = (x - a)(x - c) + 2(x - b)(x - d)$.We evaluate the function at the given points $a, b, c, d$:$f(a) = (a - a)(a - c) + 2(a - b)(a - d) = 0 + 2(a - b)(a - d)$. Since $a  0$.$f(b) = (b - a)(b - c) + 2(b - b)(b - d) = (b - a)(b - c) + 0$. Since $b > a$ and $b  0$ and $(b - c) $f(c) = (c - a)(c - c) + 2(c - b)(c - d) = 0 + 2(c - b)(c - d)$. Since $c > b$ and $c  0$ and $(c - d) $f(d) = (d - a)(d - c) + 2(d - b)(d - d) = (d - a)(d - c) + 0$. Since $d > a$ and $d > c$,$(d - a) > 0$ and $(d - c) > 0$,so $f(d) > 0$.Since $f(x)$ is a quadratic polynomial,it is continuous. Because $f(b)  0$,there exists a root in $(a, b)$. Because $f(c)  0$,there exists a root in $(c, d)$.Since the quadratic equation has two distinct real roots,the roots are real and distinct.

If $a < b < c < d$,then the roots of the equation $(x - a)(x - c) + 2(x - b)(x - d) = 0$ are

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