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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

$\frac{1}{4}$ of a number subtracted from $\frac{1}{3}$ of the same number gives $12$. The number is:

$\frac{901}{29} \times \frac{91}{301} \div \frac{51}{599} = ?$

$\left[ {8\left\{ {\left( {\frac{{21 \times \sqrt {\frac{9}{{441}}} }}{5} \text{ of } 60\% - \frac{1}{5}} \right) \times 625 + 7} \right\} \div 4} \right] = ?$

$90 \times \frac{6}{18} + 73 = ?$

If $1$ is added to the denominator of a fraction,the fraction becomes $\frac{1}{2}$. If $1$ is added to the numerator,the fraction becomes $1$. The fraction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module