If b a,then in which interval do the roots | vedclass

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(D) Let $f(x) = (x - a)(x - b) - 1$.Since $f(x)$ is a downward-opening parabola if we consider the sign,but here the coefficient of $x^2$ is $1$,so it

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One root in $(-\infty, a)$ and the other in $(b, +\infty)$

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(D) Let $f(x) = (x - a)(x - b) - 1$.Since $f(x)$ is a downward-opening parabola if we consider the sign,but here the coefficient of $x^2$ is $1$,so it is an upward-opening parabola.Calculate the values at $x = a$ and $x = b$:$f(a) = (a - a)(a - b) - 1 = -1$$f(b) = (b - a)(b - b) - 1 = -1$Since $f(a) Therefore,one root lies in $(-\infty, a)$ and the other root lies in $(b, +\infty)$.

If $b > a$,then in which interval do the roots of the equation $(x - a)(x - b) - 1 = 0$ lie?

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