Question
Rationalize the denominator : $\frac{12}{4 \sqrt{3}-\sqrt{2}}$

Answer

$\frac{12}{4 \sqrt{3}-\sqrt{2}}=\frac{12}{(4 \sqrt{3}-\sqrt{2})} \times \frac{(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3}+\sqrt{2})}$
...[Multiplying the numerator and
denominator by $(4 \sqrt{3}+\sqrt{2})]$
$= \frac{12(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3}-\sqrt{2})(4 \sqrt{3}+\sqrt{2})}$
$= \frac{12(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3})^2-(\sqrt{2})^2}$
$\ldots\left[\because(a+b)(a-b)=a^2-b^2\right]$
$= \frac{12(4 \sqrt{3}+\sqrt{2})}{(16 \times 3)-2}=\frac{12(4 \sqrt{3}+\sqrt{2})}{48-2}$
$\therefore \quad \frac{12}{4 \sqrt{3}-\sqrt{2}}=\left.\frac{6(4 \sqrt{3}+\sqrt{2})}{23}=\sqrt{2}\right)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.
Prove that $\text{ar}(\triangle\text{ABP})=\text{ar}(\text{quadrilateral ABCD}).$
Rationalize the denominator : $\frac{1}{\sqrt{3}-\sqrt{2}}$
Water in a rectangular reservoir having base $80\ m$ by $60\ m$ is $6.5\ m$ deep. In what time can the water be pumped by a pipe of which the cross-section is a square of side $20\ cm$ if the water runs through the pipe at the rate of $15\ km/hr.$
If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.
The radius and the height of a right circular cone are in the ratio $5 : 12$. If its volume is $314$ cubic meter, find the slant height and the radius. $($Use $\pi=3.14).$
Mr. Shah invested Rs. 3,20,000 in a bank at 10% compound interest . He also invested Rs. 2,40,000 in mutual funds. At market rates he got Rs. 3,05,000 after 2 years. How much did he gain ? Which of his investments was more profitable ?
In each of the following determine rational numbers a and b:$\frac{4+3\sqrt5}{4-3\sqrt5}=\text{a}+\text{b}\sqrt{5}$
An angle is equal to its supplement. Determine its measure.
In the given figure, AB || CD, $\angle\text{BAE}=65^\circ$ and $\angle\text{OEC}=20^\circ.$ Find $\angle\text{ECO}.$
What length of tarpaulin $4m$ wide will be required to make a conical tent of height 8m and base radius $6m$? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately $20cm$.
$($Use $\pi=3.14)$