First slide

Theory of equations

Question

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

Moderate

A

– 2/3
B

1/3
C

– 1/3
D

2/3

Solution

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)
and $(a^{2} - 5 a + 3) (4 α^{2}) + (3 a - 1) (2 α) + 2 = 0$ (3)
Multiplying (2) by 4 and subtracting it form (3) we get
$(3 a - 1) (2 α) + 6 = 0$
Clearly $a \neq 1 / 3 .$ Therefore, $α = - 3 / (3 a - 1)$
Putting this value in (2) we get
$\begin{array}{l} (a^{2} - 5 a + 3) (9) - (3 a - 1)^{2} (3) + 2 (3 a - 1)^{2} = 0 \\ \Rightarrow 9 a^{2} - 45 a + 27 - (9 a^{2} - 6 a + 1) = 0 \\ \Rightarrow - 39 a + 26 = 0 \\ \Rightarrow a = 2 / 3 . \end{array}$
For $a = 2 / 3$ , the equation becomes $x^{2} + 9 x + 18 = 0$ , whose roots are – 3, – 6.

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