Question
Let $$a, b, c$$ be real numbers and $$a \ne 0.$$ 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 $$\gamma $$ that always satisfies
A.
$$\gamma = \frac{1}{2}\left( {\alpha + \beta } \right)$$
B.
$$\gamma = \alpha + \frac{\beta }{2}$$
C.
$$\gamma = \alpha $$
D.
$$\alpha < \gamma < \beta $$
Answer :
$$\alpha < \gamma < \beta $$
Solution :
Let $$f\left( x \right) = {a^2}{x^2} + 2bx + 2c.$$ From the question, $${a^2}{\alpha ^2} + b\alpha + c = 0$$ and $${a^2}{\beta ^2} - b\beta - c = 0.$$
Now, $$f\left( \alpha \right) = {a^2}{\alpha ^2} + 2b\alpha + 2c = b\alpha + c = - {a^2}{\alpha ^2}$$
$$f\left( \beta \right) = {a^2}{\beta ^2} + 2b\beta + 2c = 3b\beta + 3c = 3\left( {b\beta + c} \right) = 3{a^2}{\beta ^2}.$$
Clearly, $$0 < \alpha < \beta $$
⇒ $$\alpha ,\beta $$ are real. So, $$f\left( \alpha \right) < 0,f\left( \beta \right) > 0.$$
So, $$f\left( \gamma \right) = 0$$ where $$\alpha < \gamma < \beta .$$