Find the roots of the following quadratic equ | vedclass

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Question

(A) Given equation: $4x^{2} + 20x + 23 = 0$. Divide the entire equation by $4$: $x^{2} + 5x + \frac{23}{4} = 0$. Rewrite as: $x^{2} + 5x = -\frac{23}{

Vedclass · Accepted Answer

$\frac{-5+\sqrt{2}}{2}, \frac{-5-\sqrt{2}}{2}$

Vedclass · Answer

(A) Given equation: $4x^{2} + 20x + 23 = 0$. Divide the entire equation by $4$: $x^{2} + 5x + \frac{23}{4} = 0$. Rewrite as: $x^{2} + 5x = -\frac{23}{4}$. Add the square of half the coefficient of $x$ (which is $(\frac{5}{2})^{2} = \frac{25}{4}$) to both sides: $x^{2} + 5x + \frac{25}{4} = -\frac{23}{4} + \frac{25}{4}$. This simplifies to: $(x + \frac{5}{2})^{2} = \frac{2}{4} = \frac{1}{2}$. Taking the square root on both sides: $x + \frac{5}{2} = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$. Thus,$x = -\frac{5}{2} \pm \frac{\sqrt{2}}{2} = \frac{-5 \pm \sqrt{2}}{2}$. Therefore,the roots are $\frac{-5+\sqrt{2}}{2}$ and $\frac{-5-\sqrt{2}}{2}$.

Find the roots of the following quadratic equation by the method of completing the square: $4x^{2} + 20x + 23 = 0$.

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