Question

Rationalize the denominater of a) 7+3√5 / 7-3√5
b) 5+2√3/7+4√3
c) √5+√3 /√5-√3
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Answer 1

Answer:

a) 94/4

b) 11-6√3

c) 4

Step-by-step explanation:

a) $\frac{7+3\sqrt{5} }{7-3\sqrt{5} }$ × $\frac{7+3\sqrt{5} }{7+3\sqrt{5} }$ = $\frac{(7+3\sqrt{5})^{2} }{(7)^{2}-(3\sqrt{5})^{2}}$

[(a-b)×(a+b)=a²-b²]

= $\frac{49+45}{49-45}$ = $\frac{94}{4}$

b)$\frac{5+2\sqrt{3} }{7+4\sqrt{3} }$ × $\frac{7-4\sqrt{3} }{7-4\sqrt{3}}$ = $\frac{5(7-4\sqrt{3})+2\sqrt{3} (7-4\sqrt{3}) }{(7)^{2}-(4\sqrt{3})^{2} }$

[(a-b)×(a+b)=a²-b²]

= $\frac{35-20\sqrt{3}+14\sqrt{3} -24 }{49-48}$ = $\frac{11-6\sqrt{3} }{1}$ = 11-6√3

c)$\frac{\sqrt{5}+\sqrt{3} }{\sqrt{5}-\sqrt{3} }$ × $\frac{\sqrt{5}+\sqrt{3} }{\sqrt{5}+\sqrt{3} }$ = $\frac{(\sqrt{5})^{2}+(\sqrt{3})^{2} }{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$

[(a-b)×(a+b)=a²-b²]

= $\frac{5+3}{5-3}$ = $\frac{8}{2}$ = 4