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Algebra: Polynomials Test-3

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Question 1
If for non-zero, $ \displaystyle x,{{x}^{2}}-4x-1=0,\,\,\,the\,\,\,value\,\,\,of\,\,\,{{x}^{2}}+\frac{1}{{{x}^{2}}}$
A
4
B
10
C
12
D
18
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x-\frac{1}{x} \right)}^{2}}+2\\={{\left( \frac{{{x}^{2}}-1}{x} \right)}^{2}}+2\\=16+2=18\end{array}$
Question 2
$ \displaystyle If\,\,\,x=3+\sqrt{8},\,\,\,then\,\,{{x}^{2}}+\frac{1}{{{x}^{2}}}\,\,\,is\,\,\,equal\,\,to$
A
38
B
36
C
34
D
30
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2\\x=3+\sqrt{8}\\\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{\left( 3-\sqrt{8} \right)}{\left( 3+\sqrt{8} \right)\left( 3-\sqrt{8} \right)}=3-\sqrt{8}\\{{\left( x+\frac{1}{x} \right)}^{2}}-2={{\left( 3+\sqrt{8}+3-\sqrt{8} \right)}^{2}}-2=36-2=34\end{array}$
Question 3
$ \displaystyle If\,\,\,a+\frac{1}{b}=1\,\,\,and\,\,\,b+\frac{1}{c}=1,\,\,\,then\,\,\,c+\frac{1}{a}\,\,is\,\,\,equal\,\,\,to$
A
0
B
$ \displaystyle \frac{1}{2}$
C
1
D
2
Question 3 Explanation: 
$ \displaystyle \begin{array}{l}a+\frac{1}{b}=1\Rightarrow a=1-\frac{1}{b}=\frac{b-1}{b}\\\Rightarrow \frac{1}{a}=\frac{b}{b-1}\,\,and\,\,\\b+\frac{1}{c}=1\Rightarrow \frac{1}{c}=1-b\Rightarrow c=\frac{1}{1-b}\\Therefore\,\,\,c+\frac{1}{a}=\frac{b}{b-1}+\frac{1}{1-b}=1\end{array}$
Question 4
$ \displaystyle If\,\,\,{{x}^{2}}-3x+1=0,\,\,then\,\,\,the\,\,\,value\,\,\,of\,\,\,x+\frac{1}{x}\,\,is$
A
0
B
1
C
2
D
3
Question 4 Explanation: 
$ x+\frac{1}{x}=\frac{{{x}^{2}}+1}{x}=\frac{3x}{x}=3$
Question 5
$ \displaystyle If\,\,x=\sqrt{3}+\sqrt{2},\,\,then\,\,the\,\,value\,\,of\,\,\,\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)\,\,is$
A
$ \displaystyle 6\sqrt{3}$
B
$ \displaystyle 12\sqrt{3}$
C
$ \displaystyle 18\sqrt{3}$
D
$ \displaystyle 24\sqrt{3}$
Question 5 Explanation: 
$ \displaystyle \begin{array}{l}x+\frac{1}{x}=\sqrt{3}+\sqrt{2}+\frac{1}{\sqrt{3}+\sqrt{2}}\\=\sqrt{3}+\sqrt{2}\,+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\=2\sqrt{3}\\Now\,\,we\,\,have\,\,{{\left( x+\frac{1}{x} \right)}^{3}}={{x}^{3}}+\frac{1}{{{x}^{3}}}+3.x\times \frac{1}{x}\times \left( x+\frac{1}{x} \right)\\{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( 2\sqrt{3} \right)}^{3}}-3\left( 2\sqrt{3} \right)=18\sqrt{3}\\\\\,\end{array}$
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