Question

rationalize 7 + 3 root 5 / 7 minus 3 root 5

Answer 1

hy
7+3√5/7-3√5

7+3√5/7-3√5× 7+3√5/7+3√5

49+45+42√5/49-45

94+42√5/4

2(47+21√5)/4

47+21√5/2 Ans

Hope it helps u friend!

Answer 2

Value of rationalization of $\bold{\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}}}$ is equal to $\bold{\frac{(47+21 \sqrt{5 )}}{2}}$

Solution:

To start with, rationalization we need to multiply $7+3 \sqrt{5}$  with both the numerator and denominator to remove the roots. Multiplying $7+3 \sqrt{5}$ we get

$\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}} \times \frac{7+3 \sqrt{5}}{7+3 \sqrt{5}}$

$\frac{(7+3 \sqrt{5})^{2}}{(7)^{2}-(3 \sqrt{5})^{2}}=\frac{49+42 \sqrt{5}+45}{49-45}=\frac{94+42 \sqrt{5}}{4}$

(we get the values in $(a+b)^{2}$  mode in the numerator and $\left(a^{2}-b^{2}\right)$ mode in the denominator. Therefore, the primary value after multiplication is  

Taking 2 common in both numerator and denominator we get $\frac{94+42 \sqrt{5}}{4}$

$\bold{\frac{2(47+21 \sqrt{5})}{2.2}=\frac{(47+21 \sqrt{5})}{2}}$