Question

Find the value of 'a' if one root of the quadratic equation ($a^2$ - 5a + 3)$x^2$ + (3a - 1) x + 2 = 0 is twice as large as the other.

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{-3}{2}$
• D. $\frac{-2}{3}$

Answer 1

The correct option is B

$\frac{2}{3}$

Let the roots be α, 2α. Using the results for sum of the roots and product of the roots, we will eliminate α.

α + 2α = $\frac{-(3\alpha -1)}{(a^2-5a+3)}$ [Sum of the roots]

3α = $\frac{-(3\alpha -1)}{(a^2-5a+3)}$ --------------------(i)

α × 2α = $\frac{2}{a^2-5a+3}$ [Product of the roots]

${\alpha }^2$ = $\frac{1}{a^2-5a+3}$ -----------------------(ii)

Squaring(i) gives

9${\alpha }^2$ = $\frac{{(3\alpha -1)}^2}{{(a^2-5a+3)}^2}$

⇒ ${\alpha }^2$ = $\frac{1}{9}$ $\frac{{(3\alpha -1)}^2}{{(a^2-5a+3)}^2}$ --------------------------(iii)

From (ii), (iii) we get

$\frac{1}{9}$ $\frac{{(3\alpha -1)}^2}{{a^2-5a+3)}^2}$ = $\frac{1}{a^2-5a+3}$

⇒[${\alpha }^2$ - 5a + 3 ≠ 0 since the given equation is quadratic ]

⇒ ${(3\alpha -1)}^2$ = 9 × (${\alpha }^2$ - 5a + 3)

9${\alpha }^2$ - 6a + 1 = 9${\alpha }^2$ - 45a + 27

39a = 26

a = $\frac{26}{39}$

= $\frac{2}{3}$