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(C) Given equation: $x^{2}+3x-5=0$.To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{3}{2})^{2} =

Vedclass · Accepted Answer

$\frac{-3-\sqrt{29}}{2}$ and $\frac{-3+\sqrt{29}}{2}$

Vedclass · Answer

(C) Given equation: $x^{2}+3x-5=0$.To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{3}{2})^{2} = \frac{9}{4}$.$x^{2}+3x+\frac{9}{4}-5-\frac{9}{4}=0$$(x+\frac{3}{2})^{2} - \frac{20+9}{4} = 0$$(x+\frac{3}{2})^{2} - \frac{29}{4} = 0$$(x+\frac{3}{2})^{2} = (\frac{\sqrt{29}}{2})^{2}$Taking the square root on both sides:$x+\frac{3}{2} = \pm \frac{\sqrt{29}}{2}$$x = -\frac{3}{2} \pm \frac{\sqrt{29}}{2}$$x = \frac{-3 \pm \sqrt{29}}{2}$Thus,the solutions are $\frac{-3-\sqrt{29}}{2}$ and $\frac{-3+\sqrt{29}}{2}$.

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

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