Question

answer fast please guys

Answer 1

Question :

Rationalize the denominator and find a and b -

$\mathrm{\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}b\sqrt{5}}$

Solution :

Now, $\mathrm{\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}}$

$\mathrm{=\frac{(7+\sqrt{5})(7+\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}-\frac{(7-\sqrt{5})(7-\sqrt{5})}{(7+\sqrt{5})(7-\sqrt{5})}}$

[ by multiplying both the numerator and the denominator of each term with the respective conjugates of their denominators ]

$\mathrm{=\frac{49+14\sqrt{5}+5}{49-5}-\frac{49-14\sqrt{5}+5}{49-5}}$

$\mathrm{=\frac{54+14\sqrt{5}}{44}-\frac{54-14\sqrt{5}}{5}}$

$\mathrm{=\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}}$

$\mathrm{=\frac{28\sqrt{5}}{44}}$

$\mathrm{=\frac{7}{11}\sqrt{5}}$ ,

which is the required rationalized value of the given term. (Ans. 1)

Given, $\mathrm{\frac{7}{11}\sqrt{5}=a+\frac{7}{11}b\sqrt{5}}$

Comparing among coefficients, we get a = 0 & b = 1 (Ans. 2)

Answer 2

Answer:

Step-by-step explanation: