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(A) Given equation: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$.To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(

Vedclass · Accepted Answer

$\sqrt{3}, 1$

Vedclass · Answer

(A) Given equation: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$.To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{\sqrt{3}+1}{2})^{2}$.$x^{2}-(\sqrt{3}+1) x + (\frac{\sqrt{3}+1}{2})^{2} - (\frac{\sqrt{3}+1}{2})^{2} + \sqrt{3} = 0$.$(x - \frac{\sqrt{3}+1}{2})^{2} = (\frac{\sqrt{3}+1}{2})^{2} - \sqrt{3}$.$(x - \frac{\sqrt{3}+1}{2})^{2} = \frac{3+1+2\sqrt{3}}{4} - \sqrt{3} = \frac{4+2\sqrt{3}-4\sqrt{3}}{4} = \frac{4-2\sqrt{3}}{4} = \frac{(\sqrt{3}-1)^{2}}{4}$.Taking square root on both sides: $x - \frac{\sqrt{3}+1}{2} = \pm \frac{\sqrt{3}-1}{2}$.Case $1$: $x = \frac{\sqrt{3}+1 + \sqrt{3}-1}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Case $2$: $x = \frac{\sqrt{3}+1 - (\sqrt{3}-1)}{2} = \frac{\sqrt{3}+1-\sqrt{3}+1}{2} = \frac{2}{2} = 1$.Thus,the roots are $\sqrt{3}$ and $1$.

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$

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