The value of ‘a’ for which one root of the quadratic equation (a2 − 5a + 3) x2 + (3a − 1) x + 2 = 0 is twice as large as the other, is

Question

The value of ‘a’ for which one root of the quadratic equation (a2 − 5a + 3) x2 + (3a − 1) x + 2 = 0 is twice as large as the other, is (A) (2/3) (B) -(2/3) (C) (1/3) (D) -(1/3)

Tardigrade Admin · Accepted Answer

β=2α 3α=(3a-1/a2-5a+3) ((3a-1)2/a(a2-5a+3)2)=(1/a2+5a+6) ⇒ a=(2/3).

Q. 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

$\frac{2}{3}$

$- \frac{2}{3}$

$\frac{1}{3}$

$- \frac{1}{3}$

$β = 2 α$
$3 α = \frac{3 a - 1}{a ^{2} - 5 a + 3}$
$\frac{( 3 a - 1 ) ^{2}}{a ( a ^{2} - 5 a + 3 ) ^{2}} = \frac{1}{a ^{2} + 5 a + 6}$
$\Rightarrow a = \frac{2}{3} .$

Q. The value of ‘a’ for which one root of the quadratic equation (a2−5a+3)x2+(3a−1)x+2=0 is twice as large as the other, is

32​

−32​

31​

−31​