is equal to: | Square Roots and Cube Roots Questions & Answers

Q:

If $x = 2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$ then find the value of $x^{3}-6x^{2}+6x$ ?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: B) 2

Explanation:

Given that $x = 2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$

So x - 2 = $2^{\frac{2}{3}}+2^{\frac{1}{3}}$

Now cubing on both sides, we get

$(x-2)^{3}=(2^{\frac{2}{3}}+2^{\frac{1}{3}})^{3}$

=> $x^{3}-8-6x^{2}+12x=4+2+6(x-2)$

=> $x^{3}-6x^{2}+6x=-6+8 =2$ .

Therefore, $x^{3}-6x^{2}+6x$ = 2.

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Q:

${\color{Black}If\; a= \frac{\sqrt{5}+1}{\sqrt{5}-1}\; and\; b=\frac{\sqrt{5}-1}{\sqrt{5}+1},the\; value\; of\; \left [ \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}} \right ]is:}$

A) 3/4	B) 4/3
C) 3/5	D) 5/3

Answer & Explanation Answer: B) 4/3

Explanation:

${\color{Blue}a=\frac{(\sqrt{5}+1)}{(\sqrt{5}-1)}\times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{(\sqrt{5}+1)^2}{(5-1)}=\frac{5+1+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}}$

${\color{Blue}b=\frac{(\sqrt{5}-1)}{(\sqrt{5}+1)}\times \frac{(\sqrt{5}-1)}{(\sqrt{5}-1)}=\frac{(\sqrt{5}-1)^2}{(5-1)}=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}}$

${\color{Blue}\therefore\; \; a^{2}+b^{2}=\frac{(3+\sqrt{5})^2}{4}+\frac{(3-\sqrt{5})^2}{4}=\frac{(3+\sqrt{5})^2+(3-\sqrt{5})^2}{4}=\frac{2(9+5)}{4}=7}$

${\color{Blue}Also,ab=\frac{(3+\sqrt{2})}{2}\times \frac{(3-\sqrt{2})}{2}=\frac{9-5}{4}=1}$

${\color{Blue}\therefore \; \; \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}=\frac{(a^{2}+b^{2})+ab}{(a^{2}+b^{2})-ab}=\frac{7+1}{7-1}=\frac{8}{6}=\frac{4}{3}}$

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3 594

Q:

${\color{Black}If \; x=(7-4\sqrt{3}),then\; the \; value\; of\; \left ( x+\frac{1}{x} \right )is:}$

A) 3sqrt{3}	B) 8sqrt{3}
C) 14	D) 14+8sqrt{3}

Answer & Explanation Answer: C) 14

Explanation:

${\color{Blue} x+\frac{1}{x} =(7-4\sqrt{3})+\frac{1}{(7-4\sqrt{3})}\times \frac{(7+4\sqrt{3})}{(7+4\sqrt{3})}=(7-4\sqrt{3})+\frac{(7+4\sqrt{3})}{(49-48)}}$

${\color{Blue} =(7-4\sqrt{3})+(7+4\sqrt{3})=14}$

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25 821

Q:

${\color{Black} \frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}=?}$

A) 3	B) 3sqrt{2}
C) 6	D) None of these

Answer & Explanation Answer: D) None of these

Explanation:

${\color{Blue}Given\; exp.= \frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}}$

${\color{Blue}= \frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}{3-2}=\sqrt{3}}$

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Q:

${\color{Black}If\; \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3},then:}$

A) a=-11,b=-6	B) a=-11,b=6
C) a=11,b=-6	D) a=11,b=6

Answer & Explanation Answer: C) a=11,b=-6

Explanation:

${\color{Blue} a+b\sqrt{3}=\frac{(5+2\sqrt{3})}{(7+4\sqrt{3})}\times \frac{(7-4\sqrt{3})}{(7-4\sqrt{3})}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{(7)^{2}-(4\sqrt{3})^{2}}=\frac{11-6\sqrt{3}}{49-48}=11-6\sqrt{3}}$

${\color{Blue} \therefore a=11,b=-6}$

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5 595

A) 1/6	B) 0.06
C) 0.6	D) 6

${\color{Black} \sqrt{\frac{0.204\times 42}{0.07\times 3.4}}}$ is equal to:

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