A number is greater than 58 times its recipro | vedclass

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Question

(D) Let the number be $x$.According to the problem,the number is greater than $58$ times its reciprocal by $\frac{3}{4}$.So,$x - \frac{58}{x} = \frac{

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let the number be $x$.According to the problem,the number is greater than $58$ times its reciprocal by $\frac{3}{4}$.So,$x - \frac{58}{x} = \frac{3}{4}$.Multiplying by $4x$,we get $4x^2 - 232 = 3x$.Rearranging the terms,we get the quadratic equation $4x^2 - 3x - 232 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 4, b = -3, c = -232$:$x = \frac{3 \pm \sqrt{(-3)^2 - 4(4)(-232)}}{2(4)}$$x = \frac{3 \pm \sqrt{9 + 3712}}{8}$$x = \frac{3 \pm \sqrt{3721}}{8}$$x = \frac{3 \pm 61}{8}$Taking the positive root: $x = \frac{64}{8} = 8$.Taking the negative root: $x = \frac{-58}{8} = -7.25$.Since $8$ is one of the given options,the correct number is $8$.

$A$ number is greater than $58$ times its reciprocal by $\frac{3}{4}$. What is the number?

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