Find the roots of the following quadratic equ | vedclass

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(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}$$

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$2(1+\sqrt{3})$ and $2(1-\sqrt{3})$

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(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}$$\frac{x+2+2x+2}{x^2+3x+2} = \frac{4}{x+4}$$\frac{3x+4}{x^2+3x+2} = \frac{4}{x+4}$Cross-multiplying: $(3x+4)(x+4) = 4(x^2+3x+2)$$3x^2 + 12x + 4x + 16 = 4x^2 + 12x + 8$$3x^2 + 16x + 16 = 4x^2 + 12x + 8$Rearranging terms: $x^2 - 4x - 8 = 0$Comparing with $ax^2 + bx + c = 0$,we get $a=1, b=-4, c=-8$.Discriminant $D = b^2 - 4ac = (-4)^2 - 4(1)(-8) = 16 + 32 = 48$.Since $D > 0$,real roots exist.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{-(-4) \pm \sqrt{48}}{2(1)} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$.Thus,the roots are $2(1+\sqrt{3})$ and $2(1-\sqrt{3})$.

Find the roots of the following quadratic equation using the quadratic formula,if they exist: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$; $(x \neq -1, -2, -4)$

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