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: $x - \frac{1}{x} = \frac{45}{14}$Multiply by $x$ to clear the denominator: $\frac{x^2 - 1}{x} = \frac{45}{14}$Cross-multiply: $14(

Vedclass · Accepted Answer

$\frac{7}{2}$ and $-\frac{2}{7}$

Vedclass · Answer

(C) Given equation: $x - \frac{1}{x} = \frac{45}{14}$Multiply by $x$ to clear the denominator: $\frac{x^2 - 1}{x} = \frac{45}{14}$Cross-multiply: $14(x^2 - 1) = 45x$Rearrange into standard quadratic form $ax^2 + bx + c = 0$: $14x^2 - 45x - 14 = 0$Factorize by splitting the middle term: $14x^2 - 49x + 4x - 14 = 0$Group the terms: $7x(2x - 7) + 2(2x - 7) = 0$Factor out the common binomial: $(2x - 7)(7x + 2) = 0$Set each factor to zero: $2x - 7 = 0$ or $7x + 2 = 0$Solve for $x$: $x = \frac{7}{2}$ or $x = -\frac{2}{7}$Thus,the roots of the quadratic equation are $\frac{7}{2}$ and $-\frac{2}{7}$.

Solve the following equation using the method of factorization: $x - \frac{1}{x} = \frac{45}{14} \quad (x \neq 0)$

Similar Questions

Solve the following equation using the method of factorization: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6} \,(x \neq 1, -1)$

The solution set of a quadratic equation $x^{2}+5x-14=0$ is $\ldots \ldots \ldots \ldots .$

Which constant must be added and subtracted to solve the quadratic equation $9x^{2} + \frac{3}{4}x - \sqrt{2} = 0$ by the method of completing the square?

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 roots of the quadratic equation $x^{2}-2x-15=0$ are ....... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module