First slide

Theory of equations

Question

Two non-integer roots of ${(\frac{3 x - 1}{2 x + 3})}^{4} - 5 {(\frac{3 x - 1}{2 x + 3})}^{2} + 4 = 0$ are

Easy

A

$- \frac{2}{7}, - \frac{5}{7}$
B

$- \frac{2}{5}, - \frac{5}{7}$
C

$- \frac{2}{7}, \frac{5}{7}$
D

$\frac{2}{5}, - \frac{5}{7}$

Solution

$Let {(\frac{3 x - 1}{2 x + 3})}^{2} = t$
Then, given equation becomes
$t^{2} - 5 t + 4 = 0 \Rightarrow (t - 1) (t - 4) = 0 \Rightarrow t = 1 or 4$
when, $t = 1, then \frac{3 x - 1}{2 x + 3} = \pm 1$
$\Rightarrow x = 4, \frac{- 2}{5}$
and when, t = 4then $\frac{3 x - 1}{2 x + 3} = \pm 2$
$\Rightarrow x = - 7, - 5 / 7$
Thus, two non-integer roots of given equation are $- \frac{2}{5}, - \frac{5}{7}$ ,

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