Question

7+3√5/3+√5+7-3√5/3-√5=a+√5b find the value of a and b

Answer 1

Answer:

a=3 , b=0

Step-by-step explanation:

$LHS =\frac{7+3\sqrt{5}}{3+\sqrt{5}}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}$

$=\frac{(7+3\sqrt{5})(3-\sqrt{5})+(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$=\frac{21-7\sqrt{5}+9\sqrt{5}-15+21+7\sqrt{5}-9\sqrt{5}-15}{3^{2}-(\sqrt{5})^{2}}$

$=\frac{42-30}{9-5}$

$=\frac{12}{4}\\=3\:--(1)$

$RHS = a+\sqrt{5}b \:--(2)$

/* From (1) and (2)

$3+0\times b = a+\sqrt{5}b$

/* Compare both sides, we get

a = 3 , b = 0

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Answer 2

Given:

$\bold{\frac{7+3\sqrt{5}}{3+\sqrt{5}}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b}$

To find:

a and b= ?

Solution:

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b$

Solve L.H.S part:

$\Rightarrow \frac{7+3\sqrt{5}}{3+\sqrt{5}}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}\\\\\Rightarrow \frac{(7+3\sqrt{5})(3-\sqrt{5}) + (3+\sqrt{5})(7-3\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}\\\\\Rightarrow \frac{(21-7\sqrt{5}+9\sqrt{5}-15)+(21-9\sqrt{5}+7\sqrt{5}-15)}{((3^2)-(\sqrt{5})^2)}\\\\\Rightarrow \frac{(6+2\sqrt{5})+(6-2\sqrt{5})}{9-5}\\\\\Rightarrow \frac{12}{4}\\\\ \Rightarrow 3$

if L.H.S = R.H.S  then:

$\Rightarrow 3a+ i\sqrt{b} = a+i\sqrt{b}\\\\\Rightarrow a= 3 \ \ \ and \ \ \ b= 0$