Solve the following equation using the method | vedclass

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

Question

(A) Given equation: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$Taking $LCM$ on the left side: $\frac{x(x+1) + 1(x+2)}{x(x+2)} = \frac{5}{4}$$\frac{x

Vedclass · Accepted Answer

$\{2, -4\}$

Vedclass · Answer

(A) Given equation: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$Taking $LCM$ on the left side: $\frac{x(x+1) + 1(x+2)}{x(x+2)} = \frac{5}{4}$$\frac{x^2 + x + x + 2}{x^2 + 2x} = \frac{5}{4}$$\frac{x^2 + 2x + 2}{x^2 + 2x} = \frac{5}{4}$Cross-multiplying: $4(x^2 + 2x + 2) = 5(x^2 + 2x)$$4x^2 + 8x + 8 = 5x^2 + 10x$Rearranging terms to form a quadratic equation: $x^2 + 2x - 8 = 0$Factorizing the quadratic equation: $x^2 + 4x - 2x - 8 = 0$$x(x+4) - 2(x+4) = 0$$(x-2)(x+4) = 0$Thus,$x = 2$ or $x = -4$.The solution set is $\{2, -4\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$

Similar Questions

Solve the following equation using the method of factorization and write its solution set: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{7}{2x}$ $(x \neq 2, x \neq -3, x \neq 0)$

If one of the roots of the quadratic equation $x^{2}+bx-15=0$ is $3$,then the other root is ..... .

The roots of $4x^{2} + 2x + \frac{1}{4} = 0$ are $\ldots \ldots \ldots \ldots.$

If $x=-2$ is a solution of a quadratic equation $k x^{2}+5 x+2=0,$ then $k=\ldots \ldots \ldots \ldots$

Solve the following equation using the method of factorization: $x + 3 + \frac{2}{x} = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module