Since, \( \Large \alpha \) and \( \Large \beta \) are the roots of given equation

Let \( \Large f \left(x\right)=a^{2}x^{2}+2bx+2c=0 \)

then \( \Large f \left( \alpha \right)=a^{2}a^{2}+2ba+2c=0 \)

=\( \Large a^{2}a^{2}+2 \left(b \alpha +c\right)=a^{2}a^{2}-2a^{2}a^{2} \)

=\( \Large -a^{2}a^{2}= -ve \)

and \( \Large f \left( \beta \right)=a^{2} \beta ^{2}+2 \left(b \beta +c\right)=a^{2} \beta ^{2}+2a^{2} \beta ^{2} \)

= \( \Large 3a^{2} \beta ^{2} = +ve \)

since, \( \Large f \left( \alpha \right) \) and \( \Large f \left( \beta \right) \) are of opposite signs therefore by theory of equations there lies a root \( \Large \gamma \) of the equation \( \Large f \left(x\right)=0 \) between \( \Large \alpha \) and \( \Large \beta \) i.e., a < \( \Large \gamma \) < \( \Large \beta \).