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: $x - \frac{1}{x} - \frac{45}{14} = 0$.Multiplying the entire equation by $14x$ (the least common multiple of the denominators) to

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given equation: $x - \frac{1}{x} - \frac{45}{14} = 0$.Multiplying the entire equation by $14x$ (the least common multiple of the denominators) to clear the fractions:$14x^2 - 14 - 45x = 0$.Rearranging into standard quadratic form $ax^2 + bx + c = 0$:$14x^2 - 45x - 14 = 0$.To factorize,we look for two numbers whose product is $14 	imes (-14) = -196$ and whose sum is $-45$. These numbers are $-49$ and $4$.$14x^2 - 49x + 4x - 14 = 0$.Grouping the terms:$7x(2x - 7) + 2(2x - 7) = 0$.$(2x - 7)(7x + 2) = 0$.Setting each factor to zero:$2x - 7 = 0 \implies x = \frac{7}{2}$.$7x + 2 = 0 \implies x = -\frac{2}{7}$.Thus,the solution set is $\{ \frac{7}{2}, -\frac{2}{7} \}$.

Solve the following equation using the method of factorization and write its solution set: $x - \frac{1}{x} - \frac{45}{14} = 0$; $x \neq 0$.

Similar Questions

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

The name of the Indian mathematician who gave the formula for finding the roots of the quadratic equation $ax^2 + bx + c = 0$ by the method of completing the square is ..... .

Examine whether the following equation is quadratic or not: $(x-3)(2x+1) = x(x+5)$

If $x=\sqrt{2}$ is one of the roots of the equation $ax^{2}+\sqrt{2}bx+2c=0$; $a \neq 0$,$a, b, c \in R$,then prove that $a+b+c=0$.

Solve the following equation using the method of 'completing the square': $x^{2}+3x-5=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module