For each question,select the proper option fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the rationalising factor of an expression of the form $a - \sqrt{b}$,we multiply it by its conjugate $a + \sqrt{b}$ to eliminate the squar

Vedclass · Accepted Answer

$4+\sqrt{5}$

Vedclass · Answer

(B) To find the rationalising factor of an expression of the form $a - \sqrt{b}$,we multiply it by its conjugate $a + \sqrt{b}$ to eliminate the square root.Here,the given expression is $4 - \sqrt{5}$.Multiplying $4 - \sqrt{5}$ by $4 + \sqrt{5}$ gives $(4)^2 - (\sqrt{5})^2 = 16 - 5 = 11$,which is a rational number.Therefore,the rationalising factor of $4 - \sqrt{5}$ is $4 + \sqrt{5}$.

For each question,select the proper option from four options given,to make the statement true: The rationalising factor of $4-\sqrt{5}$ is...........

Similar Questions

The number obtained on rationalizing the denominator of $\frac{1}{7-\sqrt{2}}$ is

Express $0.7\overline{39}$ in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.

Show that $0.142857142857... = 0.\overline{142857} = \frac{1}{7}$.

Prove that $\left(\frac{x^{a}}{x^{b}}\right)^{a+b} \times \left(\frac{x^{b}}{x^{c}}\right)^{b+c} \times \left(\frac{x^{c}}{x^{a}}\right)^{c+a} = 1$.

Rationalise the denominator of the following:
$\frac{16}{\sqrt{41}-5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module