For what value of $a$ , the difference of roots of the equation $(a - 2) x^{2} - (a - 4) x - 2 = 0$ is equal to $3$ .

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Solution

Given equation $(a - 2) x^{2} - (a - 4) x - 2 = 0$

$α + β = - \frac{b}{a} = \frac{a - 4}{a - 2}$ ...(1)

$α - β = 3$ ...(2) ,given

$α β = \frac{c}{a} = - \frac{2}{a - 2}$

(1)+(2) gives

$α = \frac{1}{2} (3 + \frac{a - 4}{a - 2})$

$β = \frac{1}{2} (\frac{a - 4}{a - 2} - 3)$

$α β = α β$

$\frac{1}{2} (3 + \frac{a - 4}{a - 2}) \times \frac{1}{2} (\frac{a - 4}{a - 2} - 3) = - \frac{2}{a - 2}$

$(3 a - 6 + a - 4) (a - 4 - 3 a + 6) = - 8$

$(2 a - 5) (1 - a) = - 2$

$- 2 a^{2} + 7 a - 5 = - 2$

$2 a^{2} - 7 a + 3 = 0$

$(a - 3) (2 a - 1) = 0$

$∴ a = 3, \frac{1}{2}$

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