Question

5.) if a=3 + root 5 ,then find the value of a square + 1/a square. I am series so don't add irtrravalent answer

simplify

I) (1 ki power 3 + 2 ki power 3+ 3 ki power 3) power 1/2

2) {3/4} ki power 4
{8/5} ki power -12
{32/5} ki power 6

3) (1/27) ki power-2/3

4) [{(625)ki power-1/2}ki power-1/4] ki power 2

5) 64 ki power-1/3(64 ki power 1/3- 64 ki power 2/3)

6) 8 ki power 1/3×16 ki power 1/3÷ 32 ki power -1/3

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Answer 1

Answer:

mark me as brain list

Step-by-step explanation:

Answer 2

Answer:

$\cdot \frac{3^{2}}{2^{2}}+\frac{1}{2^{2}}=9-2=7$

Step-by-step explanation:$\begin{array}{l}\begin{array}{l}a=\frac{3+\sqrt{5}}{2} \\\therefore \quad \frac{1}{a}=\frac{2}{3+\sqrt{5}}\end{array}\\\text { On rationaliging the denominator, we get, }\\\frac{1}{a}=\frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}\\\begin{aligned}&=\frac{6-2 \sqrt{5}}{9-5} \\&=\frac{6-2 \sqrt{5}}{4} \\=& \frac{3-\sqrt{5}}{2}\end{aligned}\end{array}\begin{array}{l}\text { Also, }\\\left(a+\frac{1}{a}\right)^{2}=a^{2}+\frac{1}{a^{2}}+2\end{array}$$Substituting the values of \mathrm{a} \neq \frac{1}{\mathrm{a}} ,We get, \cdot\left(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\right)^{2}=\left(a^{2}+\frac{1}{a^{2}}+2\right) \therefore \quad a^{2}+\frac{1}{a^{2}}+2=\left(\frac{3+\sqrt{5}+3-\sqrt{5}}{2}\right)^{2} =(3)^{2}=9 \therefore a^{2}+\frac{1}{a^{2}}=9-2=7$