Question

rationalising the denominator of 1/root 7-2

Answer 1

Answer:

The answer is $\sqrt{7}$ + 2/3

Step-by-step explanation:

To rationalize the denominator of the number given in the question

Step 1 = Multiply both the denominator and the numerator of 1/$\sqrt{7}$ - 2 with the conjugate of the denominator

The conjugate of the denominator($\sqrt{7}$ - 2) is

That is

=  1/$\sqrt{7}$ - 2 X

Step 2 = Follow the identity

That is

= $\sqrt{7}$ + 2[Using multiplication property]/(

= $\sqrt{7}$ + 2/7 - 4

= $\sqrt{7}$ + 2/3

Answer 2

$\textbf{ \underline{Question :} } \\ \\ </p><p></p><p> \text{Rationalise \: the \: denominator \: of }\bf{ \dfrac{1}{ \sqrt{7 } - 2} } \\ \\ \textbf{ \underline{Solution :}}\\ \\ \sf{\red{Multiplying\:both\: the\:numerator\:and\:the\:denominator\:of }}\\ \sf{\red{\bigg(\dfrac{1}{ \sqrt{7 } - 2}\bigg)\: with\:the\:conjugate\:of\:its\: denominator \big(\sqrt{7 } + 2\big)}}\\ \\ \qquad \: \bf{ \dfrac{1}{ \sqrt{7} - 2} \times \dfrac{\sqrt{7 } + 2}{\sqrt{7 } + 2} } \\ \\ \small{ \textsf{comparing\:the\:denominator\:with\:the\:alzebric\:formula :}} \\ \qquad \: \fbox{ \red{ \sf{{x}^{2} - {y}^{2} = (x + y)(x - y)}}} \\ \\\sf{Here ,}\\ \quad \: \red{ \bullet} \: \: \sf{ \small{ x = \sqrt{7}}}\\ \quad \: \red{ \bullet} \: \: \sf{ \small{ y = 2}}\\ \\ \rightarrow \fbox{ \red{\bf{\small{ { \big(\sqrt{7}} \big)^{2} - {2}^{2} = \big(\sqrt{7} + 2 \big)\big(\sqrt{7} - 2 \big)}}}} \\ \\ \longrightarrow \: \dfrac{ \sqrt{7} + 2 }{ { \big(\sqrt{7}} \big)^{2} - {2}^{2}} \\ \\ \longrightarrow \: \dfrac{ \sqrt{7} + 2 }{ 7 - 4} \\ \\ \longrightarrow \fbox{ \bf{\red{\: \dfrac{ \sqrt{7} + 2 }{ 3} } }}\\ \\ \underline{ \sqrt{7} = 2.64 \: ,\textsf{ \scriptsize{putting \: the \: value \: in \: the \: answer \: we \: got \: above}} \: \: }\\ \\ \longrightarrow \: \dfrac{ 2.64 + 2 }{ 3}\\ \\ \longrightarrow \: \dfrac{ 4.64}{ 3} \\ \\ \longrightarrow \: \red{1.546} \\ \\ \bf{so \: on \: rationalising \: the \: denominator \: of }\bf{ \dfrac{1}{ \sqrt{7 } - 2} } \\ \underline{ \bf{we \: got \: \bigg(\dfrac{ \sqrt{7} + 2 }{ 3} \bigg) \: or \: \big(1.546 \big)} \: \: \: \: \: \: \: }$