The sum of a non-zero number and twice its re | vedclass

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Question

(D) Let the number be $x$.According to the problem,the sum of the number and twice its reciprocal is $\frac{33}{4}$.So,$x + 2(\frac{1}{x}) = \frac{33}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let the number be $x$.According to the problem,the sum of the number and twice its reciprocal is $\frac{33}{4}$.So,$x + 2(\frac{1}{x}) = \frac{33}{4}$.Multiply the entire equation by $4x$ to clear the denominators:$4x^2 + 8 = 33x$.Rearrange the terms to form a quadratic equation:$4x^2 - 33x + 8 = 0$.Factor the quadratic equation:$4x^2 - 32x - x + 8 = 0$.$4x(x - 8) - 1(x - 8) = 0$.$(4x - 1)(x - 8) = 0$.This gives two possible values for $x$: $x = \frac{1}{4}$ or $x = 8$.Comparing these with the given options,the correct value is $8$.

The sum of a non-$zero$ number and twice its reciprocal is $\frac{33}{4}$. Find the number.

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