Question

If $x=\frac{1}{(2-\sqrt{3})}$, show that the value of (x³ − 2x² − 7x + 5) is 3.
Or
Simplify: $\frac{1}{\left(1+\sqrt{2}\right)}+\frac{1}{\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{\left(\sqrt{3}+\sqrt{4}\right)}+......+\frac{1}{\left(\sqrt{8}+\sqrt{9}\right)}$

Answer 1

$$\begin{gathered} x=\frac{1}{2-\sqrt{3}} \\ \mathrm{By}\mathrm{rationalising}\mathrm{we}\mathrm{get}: \\ x=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\ \Rightarrow x=\frac{2+\sqrt{3}}{4-3} \\ \Rightarrow x=2+\sqrt{3} \\ \Rightarrow x-2=\sqrt{3} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ {\left(x-2\right)}^2=3 \\ \Rightarrow x^2+4-4x=3 \\ \Rightarrow x^2-4x+1=0 \\ \mathrm{So},\mathrm{we}\mathrm{can}\mathrm{write}{x}^3-2x^2-7x+5\mathrm{as}{x}^3-4x^2+x+2x^2-8x+2+3. \\ \mathrm{Hence},x^3-2x^2-7x+5=x^3-4x^2+x+2x^2-8x+2+3 \\ =x\left(x^2-4x+1\right)+2\left(x^2-4x+1\right)+3 \\ =x\left(0\right)+2\left(0\right)+3 \\ =3 \end{gathered}$$

OR

$$\begin{gathered} \frac{1}{\left(1+\sqrt{2}\right)}+\frac{1}{\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{\left(\sqrt{3}+\sqrt{4}\right)}+......+\frac{1}{\left(\sqrt{8}+\sqrt{9}\right)} \\ \text{B}y\mathrm{rationalising}\mathrm{we}\mathrm{get}: \\ \frac{1}{\left(1+\sqrt{2}\right)}=\frac{1}{\left(\sqrt{2}+1\right)}\times \frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}=\frac{\left(\sqrt{2}-1\right)}{2-1}=\sqrt{2}-1 \\ \frac{1}{\left(\sqrt{2}+\sqrt{3}\right)}=\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)}\times \frac{\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)}=\frac{\left(\sqrt{3}-\sqrt{2}\right)}{3-2} \end{gathered}$$


$$\begin{gathered} =\sqrt{3}-\sqrt{2} \\ \mathrm{Similarly}, \\ \frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3} \\ \frac{1}{\sqrt{4}+\sqrt{5}}=\sqrt{5}-\sqrt{4}\frac{1}{\sqrt{5}+\sqrt{6}}=\sqrt{6}-\sqrt{5}\frac{1}{\sqrt{6}+\sqrt{7}}=\sqrt{7}-\sqrt{6}\frac{1}{\sqrt{7}+\sqrt{8}}=\sqrt{8}-\sqrt{7}\frac{1}{\sqrt{8}+\sqrt{9}}=\sqrt{9}-\sqrt{8}\mathrm{Adding}\mathrm{all}\mathrm{the}\mathrm{terms}\mathrm{of}\text{the}\mathrm{left}\mathrm{side}\mathrm{and}\mathrm{right}\mathrm{side},\mathrm{we}\mathrm{get}:\frac{1}{\left(1+\sqrt{2}\right)}+\frac{1}{\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{\left(\sqrt{3}+\sqrt{4}\right)}+.....+\frac{1}{\left(\sqrt{8}+\sqrt{9}\right)}=\sqrt{9}-1=3-1=2 \end{gathered}$$