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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

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

Let $α$ and $2 α$ be the roots of (1), then

$(a^{2} - 5 a + 3) α^{2} + (3 a - 1) α + 2 = 0 . . . . . . . . (2) a n d (a^{2} - 5 a + 3) (4 α^{2}) + (3 a - 1) (2 α) + 2 = 0 . . . . . . . (3)$

Multiplying (2) by 4 and subtracting it form (3) we get $(3 a - 1) (2 α) + 6 = 0$

Clearly $a \neq \frac{1}{3}$ Therefore, $α - \frac{- 3}{(3 a - 1)}$

Putting this value in (2) we get

$(a^{2} - 5 a + 3) (9) - (3 a - 1)^{2} (3) + 2 (3 a - 1)^{2} = 0 \Rightarrow 9 a^{2} - 45 a + 27 - (9 a^{2} - 6 a + 1) = 0 \Rightarrow - 39 a + 26 = 0 \Rightarrow a = \frac{2}{3} F o r a = \frac{2}{3}, t h e e q u a t i o n b e c o m e s x^{2} + 9 x + 18 = 0, w h o s e r o o t s a r e - 3, - 6$

Suggest Corrections

Similar questions

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 the other, is

a

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

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

Join BYJU'S Learning Program