Advertisements
Advertisements
Question
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
Options
\[3 + 2\sqrt{2}\]
\[\frac{1}{3 + 2\sqrt{2}}\]
\[3 - 2\sqrt{2}\]
\[\frac{3}{2} - \sqrt{2}\]
Advertisements
Solution
Given that `1/(sqrt9- sqrt8)`
We know that rationalization factor for `sqrt9 - sqrt8` is `sqrt9 + sqrt8`. We will multiply numerator and denominator of the given expression `1/(sqrt9- sqrt8)`by `sqrt9 + sqrt8`, to get
`1/(sqrt9- sqrt8) xx (sqrt9 + sqrt8)/(sqrt9 + sqrt8) = (sqrt9 + sqrt8)/ ((sqrt9)^2 - (sqrt8)^2) `
` = (sqrt9 + sqrt8) / (9-8)`
` = sqrt9 +sqrt2 sqrt4`
` = 3+2+sqrt2`
APPEARS IN
RELATED QUESTIONS
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
Prove that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Simplify:
`(0.001)^(1/3)`
Prove that:
`(2^(1/2)xx3^(1/3)xx4^(1/4))/(10^(-1/5)xx5^(3/5))div(3^(4/3)xx5^(-7/5))/(4^(-3/5)xx6)=10`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Determine `(8x)^x,`If `9^(x+2)=240+9^x`
For any positive real number x, find the value of \[\left( \frac{x^a}{x^b} \right)^{a + b} \times \left( \frac{x^b}{x^c} \right)^{b + c} \times \left( \frac{x^c}{x^a} \right)^{c + a}\].
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
