Question
Find rational numbers a and b such that: $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$
✓
Answer
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$ we have, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ $=\frac{\big(\sqrt{3}+\sqrt{2}\big)^2}{\big(\sqrt{3}\big)^2-\big(\sqrt{2}\big)^2}$ $=\frac{\big(\sqrt{3}\big)^2+2\times\sqrt{2}\times\sqrt{3}+\big(\sqrt{2}\big)^2}{3-2}$ $=\frac{3+2\sqrt{6}+2}{1}$ $=5+2\sqrt{6}$ Now, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$ $\Rightarrow\text{a}=5$ and $\text{b}=2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the value of $27x^3 + 8y^3$, if: $3x + 2y = 20$ and $\text{xy}=\frac{14}{9}$
→→
Find the volume, curved surface area and the total surface area of a cone whose height and slant height are $6\ cm$ and $10\ cm$ respectively. $\big(\text{Take}\ \pi=3.14\big)$
→A solid cylinder has a total surface area of $462\ cm^2$. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.
→The diameter of roller $1.5\ m$ long is $84\ cm$.If it takes $100$ revolutions to level a playground, find the cost of leveling this ground at the rate of $50$ paise per square meter.
→In Fig. $AB$ and $CD$ are two chords of a circle intersecting each other at point $E.$ Prove that $\angle\text{AEC}=\frac{1}{2}$ $($Angle subtended by arc $CXA$ at centre $+$ angle subtended by arc $DYB$ at the centre$).$
→Using factor theorem, factorize the following polynomials: $x^3 + 2x^2 - x - 2$
→Using rulers and compasses only, construct a $\triangle\text{ABC}$ from the following data:
$AB + BC + CA = 12\ cm$, $\angle\text{B}=45^\circ$ and $\angle\text{C}=60^\circ$
→$AB + BC + CA = 12\ cm$, $\angle\text{B}=45^\circ$ and $\angle\text{C}=60^\circ$
Simplify $\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$
→Express $2.\overline{36}+0.\overline{23}$ as a fraction in simplest form.
→