Solve the following equation using the quadra | vedclass

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

Question

(C) Given equation: $\frac{1}{x} - \frac{1}{x-2} = 3$. Taking $LCM$: $\frac{(x-2) - x}{x(x-2)} = 3$. $\frac{-2}{x^2 - 2x} = 3$. $-2 = 3x^2 - 6x$. $3x^

Vedclass · Accepted Answer

$\frac{3+\sqrt{3}}{3}, \frac{3-\sqrt{3}}{3}$

Vedclass · Answer

(C) Given equation: $\frac{1}{x} - \frac{1}{x-2} = 3$. Taking $LCM$: $\frac{(x-2) - x}{x(x-2)} = 3$. $\frac{-2}{x^2 - 2x} = 3$. $-2 = 3x^2 - 6x$. $3x^2 - 6x + 2 = 0$. Comparing with $ax^2 + bx + c = 0$,we get $a = 3, b = -6, c = 2$. Discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$. Since $D > 0$,real roots exist. Using quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$: $x = \frac{-(-6) \pm \sqrt{12}}{2(3)} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $\frac{1}{x} - \frac{1}{x-2} = 3, x \neq 0, 2$.

Similar Questions

Check whether the following is a quadratic equation:
$(x+2)^{3} = x^{3}-4$

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $6x^{2}-5x-1=0$.

........ also mentioned a method to find the solution of quadratic equation.

Examine whether the following equation is quadratic or not: $x(x+1)+8=(x+2)(x-2)$

Check whether the following is a quadratic equation:
$x^{2}+3x+1=(x-2)^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module