Let $a,\ b,c$ be real number $a
eq 0.If\ \alpha$ is a root of $a^{2}x^{2} + bx + c = 0,\ \beta$ is a root of $a^{2}x^{2} - bx - c = 0$ and $0

Question

Let $a,\ b,c$ be real number $a 
eq 0.If\ \alpha$ is a root of $a^{2}x^{2} + bx + c = 0,\ \beta$ is a root of $a^{2}x^{2} - bx - c = 0$ and $0 < \alpha < \beta$, then the equation $a^{2}x^{2} + 2bx + 2c = 0$ has a root of $\gamma$ that always satisfies

Dr. Abubacker · Accepted Answer

(d)
Since, $\alpha\ and\ \beta$ are the roots of given equation.
Let $f(x) = a^{2}x^{2} + 2bx + 2c = 0$
Then, $f(\alpha) = a^{2}\alpha^{2} + 2b\alpha + 2c = 0$
$= a^{2}\alpha^{2} + 2(b\alpha + c) = a^{2}\alpha^{2} - 2a^{2}\alpha^{2}$
$= - a^{2}\alpha^{2} = - ve$
and $f(\beta) = a^{2}\beta^{2} + 2(b\beta + c) = a^{2}\beta^{2} + 2a^{2}\beta^{2}$
$= 3a^{2}\beta^{2} = +$ve
Since, $f(\alpha)$ and $f(\beta)$ are of opposite signs therefore by theory of equations there lies a root $\gamma$ of the equation $f(x) = 0$ between $\alpha\ and\ \beta,\ ie,\ \alpha < \gamma < \beta$.

Q. Let $a,\ b,c$ be real number $a \neq 0.If\ \alpha$ is a root of $a^{2}x^{2} + bx + c = 0,\ \beta$ is a root of $a^{2}x^{2} - bx - c = 0$ and $0 < \alpha < \beta$, then the equation $a^{2}x^{2} + 2bx + 2c = 0$ has a root of $\gamma$ that always satisfies

Explanation

Reviewed by Dr. Abubacker

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