If the roots of the following quadratic equat | vedclass

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Question

(A) Given equation: $(2x + 1) - \frac{4}{(2x + 1)} - 3 = 0$.Let $y = 2x + 1$. Then the equation becomes $y - \frac{4}{y} - 3 = 0$.Multiplying by $y$ (

Vedclass · Accepted Answer

$\frac{3}{2}, -1$

Vedclass · Answer

(A) Given equation: $(2x + 1) - \frac{4}{(2x + 1)} - 3 = 0$.Let $y = 2x + 1$. Then the equation becomes $y - \frac{4}{y} - 3 = 0$.Multiplying by $y$ (where $y 
eq 0$): $y^2 - 3y - 4 = 0$.To solve by completing the square: $y^2 - 3y = 4$.Add $(\frac{3}{2})^2 = \frac{9}{4}$ to both sides: $y^2 - 3y + \frac{9}{4} = 4 + \frac{9}{4}$.$(y - \frac{3}{2})^2 = \frac{16 + 9}{4} = \frac{25}{4}$.Taking the square root: $y - \frac{3}{2} = \pm \frac{5}{2}$.Case $1$: $y = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4$. Since $y = 2x + 1$,$2x + 1 = 4 \implies 2x = 3 \implies x = \frac{3}{2}$.Case $2$: $y = \frac{3}{2} - \frac{5}{2} = -\frac{2}{2} = -1$. Since $y = 2x + 1$,$2x + 1 = -1 \implies 2x = -2 \implies x = -1$.The roots are $\frac{3}{2}, -1$.

If the roots of the following quadratic equation exist,find them by the method of completing the square: $(2x + 1) - \frac{4}{(2x + 1)} - 3 = 0$.

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