Question

One factor of $-x^2+x\sqrt{3}+6$ is __________.

• A. $2\sqrt{3}+x$
• B. $-2\sqrt{3}+\sqrt{3}x$
• C. $x-\sqrt{3}$
• D. $2\sqrt{3}-x$

Answer 1

Answer: (D)

$2\sqrt{3}-x$

Given : $-x^2+x\sqrt{3}+6$

Writing this as,

$-(x^2-x\sqrt{3})+6$

$\Rightarrow -(x^2-x\sqrt{3}+\frac{3}{4})+\frac{3}{4}+6$

$\Rightarrow -(x-\frac{\sqrt{3}}{2}{)}^2+\frac{27}{4}$

$\because a^2-2ab+b^2=(a-b{)}^2)$

$\Rightarrow (\frac{3\sqrt{3}}{2}{)}^2-(x-\frac{\sqrt{3}}{2}{)}^2$

$\Rightarrow (\frac{3\sqrt{3}}{2}+x-\frac{\sqrt{3}}{2})(\frac{3\sqrt{3}}{2}-x+\frac{\sqrt{3}}{2})$ $\because a^2-b^2=(a+b)(a-b)$

$\therefore$ Factors are $(x+\sqrt{3})$ and $(2\sqrt{3}-x)$