Find the value of $a$ for which one root of the equation $x^{2} + (2 a - 1) x + a^{2} + 2 = 0$ is twice as large as the other.

Open in App

Solution

$x^{2} + (2 a - 1) x + a^{2} + 2 = 0$

$l e t o n e r o o t b e α$

$o t h e r r o o t = 2 α$

$s u m o f r o o t s$

$3 α = 1 - 2 a$

$α = \frac{1 - 2 a}{3}$

$p r o d u c t o f r o o t s$

$2 α^{2} = a^{2} + 2$

$2 {(\frac{1 - 2 a}{3})}^{2} = a^{2} + 2$

$2 (4 a^{2} - 4 a + 1) = 9 a^{2} + 18$

$a^{2} + 8 a + 16 = 0$

${(a + 4)}^{2} = 0$

$∴ a = - 4$

Suggest Corrections

Similar questions

a

(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0

p

(p^{2} - 5 p + 3) x^{2} + (3 p - 1) x + 2 = 0

(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0

3 a

(a^{2} - 5 a + 3) x^{2} - 3 (a - 1) x + 2 = 0

The value of a for which one root of the quadratic equation. $(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0$ is twice as large as other, is

Join BYJU'S Learning Program