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Algebra: Quadratic Equations Test-1

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Question 1
$ \begin{array}{l}If\,\,p=99,\,\\then\,\,value\,\,of\,\,\,p\left( {{p}^{2}}+3p+3 \right)is\end{array}$
A
999
B
9999
C
99999
D
999999
Question 1 Explanation:
$ \begin{array}{l}p\left( {{p}^{2}}+3p+3 \right)\\=99\left( {{99}^{2}}+3X99+3 \right)\\=99\left( 9801+297+3 \right)\\=99(9801+300)\\=99(10101)\\=999999\end{array}$
Question 2
If p= 999,
then the value of
$ \displaystyle \sqrt[3]{p\left( {{p}^{2}}+3p+3 \right)+1}$ is
A
1000
B
999
C
998
D
1002
Question 2 Explanation:
$ \sqrt[3]{p\left( {{p}^{2}}+3p+3 \right)+1}$
$ =\sqrt[3]{999\left( 1001001 \right)+1}$
$ \displaystyle \begin{array}{l}=\sqrt[3]{999999999+1}\\=\sqrt[3]{1000000000}\\=1000\end{array}$
Question 3
If p=101,
then the value of
$ \displaystyle \sqrt[3]{p\left( {{p}^{2}}-3p+3 \right)-1}$ is
A
100
B
101
C
102
D
1000
Question 3 Explanation:
$ \sqrt[3]{p\left( {{p}^{2}}-3p+3 \right)-1}$
$ \displaystyle \begin{array}{l}\sqrt[3]{101\left( {{101}^{2}}-3X101+3 \right)-1}\\=\sqrt[3]{101\left( 10201-300 \right)-1}\\=\sqrt[3]{1000000}\\=100\end{array}$
Question 4
If p=124,
$latex \displaystyle \sqrt[3]{p\left( {{p}^{2}}+3p+3 \right)+1}$=?
A
5
B
7
C
123
D
125
Question 4 Explanation:
$ \sqrt[3]{p\left( {{p}^{2}}+3p+3 \right)+1}$
$ \displaystyle \begin{array}{l}\sqrt[3]{{{p}^{3}}+3{{p}^{2}}+3p+1}\\=p+1\\=125\end{array}$
Question 5
$ \displaystyle \begin{array}{l}If\,x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}\\Then\,\,\,5{{x}^{2}}-5x-1=?\end{array}$
A
0
B
3
C
4
D
5
Question 5 Explanation:
$ \begin{array}{l}x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}\\{{x}^{2}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}\\=>{{x}^{2}}=\frac{{{(\sqrt{5}+1)}^{2}}}{4}\\=>4{{x}^{2}}={{(\sqrt{5}+1)}^{2}}=6+2\sqrt{5}\\=>{{x}^{2}}=\frac{6+2\sqrt{5}}{4}\\5{{x}^{2}}-5x-1\\=\frac{30+10\sqrt{5}-10\sqrt{5}-10-4}{4}\\=\frac{16}{4}\\=4\end{array}$
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