alpha is a root of the equation frac{x-1}{sqr | vedclass

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(B) Given equation: $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$.Squaring both sides:$\frac{x^2-2x+1}{2x^2-5x+2} = \frac{1681}{3600}$.Cross-multiply

Vedclass · Accepted Answer

$-\frac{7}{34}$

Vedclass · Answer

(B) Given equation: $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$.Squaring both sides:$\frac{x^2-2x+1}{2x^2-5x+2} = \frac{1681}{3600}$.Cross-multiplying:$3600(x^2-2x+1) = 1681(2x^2-5x+2)$.$3600x^2 - 7200x + 3600 = 3362x^2 - 8405x + 3362$.$238x^2 + 1205x + 238 = 0$.Factoring the quadratic equation:$(34x + 7)(7x + 34) = 0$.Thus,$x = -\frac{7}{34}$ or $x = -\frac{34}{7}$.Since $-\frac{1}{2} < \alpha < 0$ and $-\frac{7}{34} \approx -0.205$ while $-\frac{34}{7} \approx -4.857$,the root $\alpha = -\frac{7}{34}$ satisfies the condition.

$\alpha$ is a root of the equation $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$. If $-\frac{1}{2} < \alpha < 0$,then $\alpha = $

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