Question
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to :
✓
Answer
- $3+2\sqrt{2}$
Solution:
After rationalising
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{\sqrt{9}-\sqrt{8}}\times\frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}$
$=\frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}$
$=\frac{\sqrt{3\times3}+\sqrt{2\times2\times2}}{9-8}$
$=\frac{3+2\sqrt{2}}{1}$
$=3+2\sqrt{2}$
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
If, for the set of observations 4, 7, x, 8, 9, 10 the mean is 8, then x is equal to:
The value of $\frac{2^\circ+7^\circ}{5^\circ}$ is:
→The point (7, 0) lies:
The decimal representation of an irrational number is:
A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is:
In figure, if l || m, then x =
- 105°
- 65°
- 40°
- 25°
The value of $\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$ is:
→Write the correct answer in the following:
After rationalising the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}},$ we get the denominator as
→After rationalising the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}},$ we get the denominator as
If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a:
The Possible expressions for the length and breadth of the rectangle whose area is given by $4a^2 + 4a - 3$ is:
→