Ex.1.5 Q5 Number System Solution - NCERT Maths Class 9

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Question

Rationalize the denominators of the following:

(i) \(\begin{align}\frac{1}{{\sqrt 7 }} \end{align}\) (ii)\(\begin{align}\frac{1}{{\sqrt 7 - \sqrt 6 }}\end{align}\)
(iii) \(\begin{align}\frac{1}{{\sqrt 5 + \sqrt 2 }}\end{align}\) (iv) \(\begin{align}\frac{1}{{\sqrt 7 - 2}}\end{align}\)

 

 Video Solution
Number Systems
Ex 1.5 | Question 5

Text Solution

 

Steps:

(i) \(\begin{align}\frac{1}{{\sqrt 7 }} = \frac{1}{{\sqrt 7 }} \times \frac{{\sqrt 7 }}{{\sqrt 7 }}\end{align}\)

(Dividing and multiplying by \(\sqrt{7}\) )

\[=\frac{\sqrt{7}}{7}\]

(ii) \(\begin{align}\frac{1}{{\sqrt 7 - \sqrt 6 }}\end{align}\)

Steps:

Dividing and multiplying by \(\begin{align}\sqrt 7 + \sqrt 6 \end{align}\), we get

\[\begin{align} &=\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}} \\ &=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}} \qquad \qquad \left(\text { Using }(a+b)(a-b)=a^{2}-b^{2}\right) \\ &=\frac{\sqrt{7}+\sqrt{6}}{7-6} \\ &=\sqrt{7}+\sqrt{6} \end{align}\]

(iii) \(\begin{align}\frac{1}{{\sqrt 5 + \sqrt 2 }}\end{align}\)

Steps:

Dividing and multiplying by \(\begin{align}\sqrt 5 - \sqrt 2 \end{align}\) , we get

\[\begin{align} &=\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ &=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}} \qquad \qquad \text{Using }(a+b)(a-b)=\left(a^{2}-b^{2}\right)\\ &=\frac{\sqrt{5}-\sqrt{2}}{5-2} \\ &=\frac{\sqrt{5}-\sqrt{2}}{3} \end{align}\]

(iv) \(\begin{align}\frac{1}{{\sqrt 7 - 2}} & \end{align}\)

Steps:

Dividing and multiplying by \(\begin{align}\sqrt 7 + 2,\end{align}\), we get

\[\begin{align} &=\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2} \\ &=\frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}} \qquad \qquad \left(\text{Using }(a+b)(a-b)={a}^{2}-{b}^{2}\right)\\ &=\frac{\sqrt{7}+2}{7-4} \\ &=\frac{\sqrt{7}+2}{3} \end{align}\]

  
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