81
Zeroes of the quadratic polynomial $p(x) = (a^2 + 10)x^2 - 74x + 7a$ are reciprocal of each other and they are rational. Find the value of 'a'.
Show SolutionHide Solution↓
Since zeroes are reciprocal of each other.
$\frac{7a}{a^2+10} = 1$
$\Rightarrow a^2 - 7a + 10 = 0$
$\Rightarrow (a-2)(a-5) = 0$
$\Rightarrow a = 2, 5$
For $a = 5$, zeroes are rational.
$\frac{7a}{a^2+10} = 1$
$\Rightarrow a^2 - 7a + 10 = 0$
$\Rightarrow (a-2)(a-5) = 0$
$\Rightarrow a = 2, 5$
For $a = 5$, zeroes are rational.