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

Vedclass · Accepted Answer

$2+2\sqrt{3}, 2-2\sqrt{3}$

Vedclass · Answer

(C) Given equation: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$.Taking $LCM$ on the left side: $\frac{(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{4}{x+4}$.Simplify the numerator: $\frac{x+2+2x+2}{x^2+3x+2} = \frac{4}{x+4} \implies \frac{3x+4}{x^2+3x+2} = \frac{4}{x+4}$.Cross-multiply: $(3x+4)(x+4) = 4(x^2+3x+2)$.Expand both sides: $3x^2 + 12x + 4x + 16 = 4x^2 + 12x + 8$.Rearrange to form a standard quadratic equation $ax^2+bx+c=0$: $x^2 - 4x - 8 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-4, c=-8$.$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-8)}}{2(1)} = \frac{4 \pm \sqrt{16 + 32}}{2} = \frac{4 \pm \sqrt{48}}{2}$.Since $\sqrt{48} = \sqrt{16 	imes 3} = 4\sqrt{3}$,we get $x = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$.Thus,the solutions are $2+2\sqrt{3}$ and $2-2\sqrt{3}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}; \, x \neq -1, -2, -4$.

Similar Questions

Obtain the roots of the following equation using the method of 'completing the square': $m^{2} - 18m + 81 = 0$

If $64$ is the discriminant of $k x^{2}-4 x-4=0$,then the value of $k$ is ...........

The product of Virat Kohli's age $8$ years ago and $6$ years later is $680$. Find his present age.

The roots of the quadratic equation $x^{2}+18x+81=0$ are ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module