Solve the following equation using the method | vedclass

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

Question

(D) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{5}{2}$.Let $y = \frac{x}{x+1}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.M

Vedclass · Accepted Answer

$\{1, -2\}$

Vedclass · Answer

(D) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{5}{2}$.Let $y = \frac{x}{x+1}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.Multiplying by $2y$,we get $2y^2 + 2 = 5y$,which simplifies to $2y^2 - 5y + 2 = 0$.Factoring the quadratic: $2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.So,$(2y - 1)(y - 2) = 0$,giving $y = 1/2$ or $y = 2$.Case $1$: $\frac{x}{x+1} = \frac{1}{2} \implies 2x = x + 1 \implies x = 1$.Case $2$: $\frac{x}{x+1} = 2 \implies x = 2x + 2 \implies x = -2$.The solution set is $\{1, -2\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{5}{2}$

Similar Questions

Solve the following equation using the method of factorization and write its solution set: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$

The value of the discriminant of the quadratic equation $5x^{2} + x + 2 = 0$ is ..... .

While selling a calculator for Rs. $56$,the profit percentage is equal to its cost price in rupees. Find the cost price of the calculator.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $k x^{2} - 24 x + 16 = 0$.

The faster train takes $1$ hour less than the slower train for covering a $400\, km$ distance. If the speed of the slower train is $20\, km/hr$ less than the speed of the faster train,find the speed of the slower train.

Vedclass Test Series

Exam Paper Generator

Online Exam Module