First slide

Theory of equations

Question

If x is real, then $\frac{x}{(x^{2} - 5 x + 9)}$ lies between

Moderate

A

-1 and -1/11
B

1 and -1/11
C

1 and 1/11
D

none of these

Solution

Let $\frac{x}{x^{2} - 5 x + 9} = y$
or ${yx}^{2} - 5 yx + 9 y = x$
or ${yx}^{2} - (5 y + 1) x + 9 y = 0$
Now, x is real, so
$D \geq 0$
$\Rightarrow (- (5 y + 1))^{2} - 4 \cdot y \cdot (9 y) \geq 0$
or $- 11 y^{2} + 10 y + 1 \geq 0$
or $11 y^{2} - 10 y - 1 \leq 0$
or $(11 y + 1) (y - 1) \leq 0$
or $- \frac{1}{11} \leq y \leq 1$

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