Let a , b ∈ R be such that the equation a x 2 - 2 b x + 15 = 0 has repeated root α and if α…
Let be such that the equation has repeated root and if and are the roots of the equation , then is equal to:
Solution
Given $ax^2 - 2bx + 15 = 0$ ... i
Has repeated roots. So $D = 0$
$4b^2 - 4 \times 15 \times a = 0$
$\Rightarrow b^2 = 15a$ ... ii
Also given
Now $\alpha$ will satisfy both quadratic
$ax^2 - 2bx + 15 = 0$ and $x^2 - 2bx + 21 = 0$
Putting the value we get
$a\alpha^2 - 2b\alpha + 15 = 0$
$\alpha^2 - 2b\alpha + 21 = 0$
$\Rightarrow (a - 1)\alpha^2 = 6$
$\alpha^2 = \frac{6}{a - 1}$
Now in equation (1) product of Root $\alpha^2 = \frac{15}{a}$
So $\frac{15}{a} = \frac{6}{a - 1}$
$\Rightarrow 2a = 5a - 5 \Rightarrow a = \frac{5}{3}$
Now $b^2 = 15a \Rightarrow b^2 = 15 \times \frac{5}{3}$
$\Rightarrow b^2 = 25$
So $b = \pm 5$
Now in quadratic $x^2 - 2bx + 21 = 0$
Putting the value of $b$ we get
$x^2 - 10x + 21 = 0 \Rightarrow (x - 7)(x - 3) = 0$
So $x = 3$ or $7$.
Or
$x^2 + 10x + 21 = 0 \Rightarrow x = -3$ or $x = -7$
So $\alpha = \pm 3$ and $\beta = \pm 7$
So $\alpha^2 + \beta^2 = 3^2 + 7^2 = 9 + 49 = 58$
