Question

someone help me plssssss it's my exams soon pls ​

Answer 1

Answer:

Hey, to rationalise you need to multiply and divide by the denominator (but with opp. sign). For instance, in this example, we'll multiply with root 15 + root 3.

So, on the numerator, we'd get : (root 15 + root 3)^2 and the denominator would be (root 15)^2 - (root 3)^3

here we used the identity : (a+b)(a-b) = a^2 - b^2

So, the answer will be: (15 + 3 + 2 root 15 × root 3)/15 - 3 = 18 + 2root60/12 = 2(9 + root60)/12

= 9 + root60/6

For further simplification, you may even solve the root.

Answer 2

$\large{\underline{\underline{\mathfrak{\green{\sf{Question:-}}}}}}$.

$\red{\:Rationalizing\:the\:Denominator\:of\:\:\frac{\sqrt{15}+\sqrt{3}}{\sqrt{15}-\sqrt{3}}}$.

$\large{\underline{\underline{\mathfrak{\pink{\sf{Solution:-}}}}}}$.

$\implies\frac{(\sqrt{15}+\sqrt{3})}{(\sqrt{15}-\sqrt{3})}$.

$\bold{\red{\:Rationalizing\:Denominator}}$

$\implies\frac{(\sqrt{15}+\sqrt{3})(\sqrt{15}+\sqrt{3}}{(\sqrt{15}+\sqrt{3})(\sqrt{15}-\sqrt{3})}$.

We know that

★ $\red{\:(a+b)^2\:=\:(a^2+b^2+2ab)}$

★$\red{\:(a^2-b^2)\:=(a+b)(a-b)}$.

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Again,

$\implies\frac{(\sqrt{15}^2+\sqrt{3}^2+2\sqrt{15}*\sqrt{3})}{(\sqrt{15}^2-\sqrt{3}^2)}$.

$\implies\frac{(\:15+3+2\sqrt{45})}{(\:15-3)}$.

$\implies\frac{(\:18+6\sqrt{5})}{12}$.

$\implies\frac{\:6(3+\sqrt{5})}{12}$.

$\implies\frac{\:(3+\sqrt{5})}{2}$.

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