Solve the following equation using the method | vedclass

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

Question

(C) Given equation: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$Let $y = \frac{x}{x-1}$,then the equation becomes $y + \frac{1}{y} = \frac{25}{12}$

Vedclass · Accepted Answer

$\{4, -3\}$

Vedclass · Answer

(C) Given equation: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$Let $y = \frac{x}{x-1}$,then the equation becomes $y + \frac{1}{y} = \frac{25}{12}$.Multiplying by $12y$,we get $12y^2 + 12 = 25y$,which rearranges to $12y^2 - 25y + 12 = 0$.Factoring the quadratic: $12y^2 - 16y - 9y + 12 = 0 \implies 4y(3y - 4) - 3(3y - 4) = 0$.So,$(4y - 3)(3y - 4) = 0$,giving $y = \frac{3}{4}$ or $y = \frac{4}{3}$.Case $1$: $\frac{x}{x-1} = \frac{3}{4} \implies 4x = 3x - 3 \implies x = -3$.Case $2$: $\frac{x}{x-1} = \frac{4}{3} \implies 3x = 4x - 4 \implies x = 4$.Thus,the solution set is $\{4, -3\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$

Similar Questions

State whether the quadratic equation $(x-1)(x+2)+2=0$ has two distinct real roots. Justify your answer.

Find the roots of the quadratic equation using the quadratic formula:
$2x^{2} - 3x - 5 = 0$

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $\sqrt{2} x^{2} + 7x + 5\sqrt{2} = 0$

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{1-x} + \frac{1-x}{x} = \frac{13}{6}$

Find the roots of the following quadratic equation by the method of completing the square: $2x^{2} - 7x + 3 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module