Question

If $x=\frac{1}{2+\sqrt{3}}$, find the value of $x^3-x^2-11x+3$

• A. $0$
• B. $3$
• C. $3x$
• D. $4$

Answer 1

The correct option is A

$0$

Step 1. Rationalizing denominator of X

We have, $x=\frac{1}{\left(2+\sqrt{3}\right)}$ we have to find the value of $x^3-x^2-11x+3$………..(1)

$$\begin{gathered} \Rightarrow x=\frac{1}{\left(2+\sqrt{3}\right)}\times \frac{\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)}(Rationalinzingthedenominatorbymultiplying2-\sqrt{3}bothnumeratoranddenominator) \\ \Rightarrow x=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)} \\ \Rightarrow x=\frac{2-\sqrt{3}}{4-3}[\because (a+b\left)\right(a-b)=a^2-b^2] \\ \Rightarrow x=2-\sqrt{3} \end{gathered}$$

Step 2. Calculating $\left(x^3-x^2-11x+3\right)$

Now $x-2=-\sqrt{3}$ squaring both sides; we get

$$\begin{gathered} \Rightarrow {\left(x-2\right)}^2={\left(-\sqrt{3}\right)}^2 \\ \Rightarrow x^2+4-4x=3 \\ \Rightarrow x^2-4x+1=0 \end{gathered}$$

Now

$$\begin{gathered} x^3-x^2-11x+3 \\ =x^3-4x^2+x+3x^2-12x+3(addingandsubstract3x^{2}intihsequation) \\ =x\left(x^2-4x+1\right)+3\left(x^2-4x+1\right)=x\times 0+3\left(0\right) \\ =0+0=0 \end{gathered}$$

Final answer : The value of $\left(x^3-x^2-11x+3\right)$ is $0$.