Find the zeros of the following quadratic pol | vedclass

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(A) To find the zeros of $p(x)$,we set $p(x) = 0$.$\sqrt{3} x^{2} - 8 x + 4 \sqrt{3} = 0$We split the middle term $-8x$ into $-6x - 2x$ because $(\sqr

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$\frac{2}{\sqrt{3}}$,$2 \sqrt{3}$

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(A) To find the zeros of $p(x)$,we set $p(x) = 0$.$\sqrt{3} x^{2} - 8 x + 4 \sqrt{3} = 0$We split the middle term $-8x$ into $-6x - 2x$ because $(\sqrt{3}) 	imes (4\sqrt{3}) = 12$,and we need two numbers whose product is $12$ and sum is $-8$.$\sqrt{3} x^{2} - 6x - 2x + 4 \sqrt{3} = 0$$\sqrt{3} x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$$(x - 2\sqrt{3})(\sqrt{3} x - 2) = 0$Setting each factor to zero:$x - 2\sqrt{3} = 0 \implies x = 2\sqrt{3}$$\sqrt{3} x - 2 = 0 \implies x = \frac{2}{\sqrt{3}}$Thus,the zeros are $\frac{2}{\sqrt{3}}$ and $2\sqrt{3}$.

Find the zeros of the following quadratic polynomial: $p(x)=\sqrt{3} x^{2}-8 x+4 \sqrt{3}$

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