Question : One of the factors of the expression $4\sqrt{3}x^{2}+5x-2\sqrt{3}$ is:
Option 1: $4x+\sqrt{3}$
Option 2: $4x+3$
Option 3: $4x-3$
Option 4: $4x-\sqrt{3}$
Correct Answer: $4x-\sqrt{3}$
Solution : $4\sqrt{3}x^{2}+5x-2\sqrt{3}$ = $4\sqrt{3}x^{2}+8x-3x-2\sqrt{3}$ = $4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)$ = $(4x-\sqrt{3})(\sqrt{3}x+2)$ Hence, the correct answer is $4x-\sqrt{3}$.
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Question : If $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$, then the value of $(a^{2}-ax)$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
Question : If $x^2-\sqrt{7} x+1=0$, then $\left(x^3+x^{-3}\right)=?$
Option 1: $4 \sqrt{7}$
Option 2: $3 \sqrt{7}$
Option 3: $10 \sqrt{7}$
Option 4: $7 \sqrt{7}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : What is the value of $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$?
Option 1: $1$
Option 2: $\frac{(x-3)}{(x+3)}$
Option 3: $\frac{(x+4)}{(x-3)}$
Option 4: $\frac{(x-3)}{(x+4)}$
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