Practice MCQ Questions and Answer on Algebra

Solution:

 $$\frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }}$$ (By rationalization of denominator) $$\eqalign{ & = \frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }} \times \frac{{7 - 4\sqrt 3 }}{{7 - 4\sqrt 3 }} \cr & = \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}}{{49 - 48}} \cr} $$

Solution:

 $$\eqalign{ & {a^2} - 4a - 1 = 0 \cr & {a^2} - 1 = 4a \cr & a - \frac{1}{a} = 4 \cr & {\text{Squring both sides}} \cr & {a^2} + \frac{1}{{{a^2}}} - 2 = 16 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = 18 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} + 3a - \frac{3}{a} \cr & = {a^2} + \frac{1}{{{a^2}}} + 3\left( {a - \frac{1}{a}} \right) \cr & = 18 + 3 \times 4 \cr & = 18 + 12 \cr & = 30 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 8 \cr & {x^2} + 1 = 8x \cr & \frac{{5x}}{{{x^2} + 1 - 6x}} \cr & = \frac{{5x}}{{8x - 6x}} \cr & = \frac{{5x}}{{2x}} \cr & = \frac{5}{2} \cr & = 2.5 \cr} $$

Solution:

 $$\eqalign{ & c + \frac{1}{c} = \sqrt 3 \cr & {\text{On cubing both side}} \cr & \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr} $$

Solution:

 $$\eqalign{ & \sqrt x + \frac{1}{{\sqrt x }} = 2\sqrt 2 \cr & x + \frac{1}{x} + 2 = 8 \cr & x + \frac{1}{x} = 6 \cr & {x^2} + \frac{1}{{{x^2}}} + 2 = 36 \cr & {x^2} + \frac{1}{{{x^2}}} = 34 \cr} $$

Solution:

 $$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \frac{1}{x} = 3 - 2\sqrt 2 \cr & x + \frac{1}{x} = 6 \cr & {x^2} + \frac{1}{{{x^2}}} = {\left( 6 \right)^2} - 2 \cr & {x^2} + \frac{1}{{{x^2}}} = 34 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{{16x}} = 3 \cr & 2x + \frac{1}{{8x}} = 6 \cr & {\text{Cube both side}} \cr & 8{x^3} + \frac{1}{{512{x^3}}} + 3 \times 2 \times \frac{1}{8} \times 6 = 216 \cr & 8{x^3} + \frac{1}{{512{x^3}}} = 216 - \frac{9}{2} \cr & {\text{Multiply by '2' both side}} \cr & 16{x^3} + \frac{1}{{256{x^3}}} = 432 - 9 = 423 \cr} $$

Solution:

 $$\eqalign{ & \frac{{\sqrt {3 + x} + \sqrt {3 - x} }}{{\sqrt {3 + x} - \sqrt {3 - x} }} = \frac{2}{1}{\text{ }} \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1} \cr & \left[ {\frac{{\text{A}}}{{\text{B}}} = \frac{{\text{C}}}{{\text{D}}}} \right] \cr & \left[ {\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = \frac{{{\text{C}} + {\text{D}}}}{{{\text{C}} - {\text{D}}}}} \right] \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = 3{\text{ }} \cr & {\text{Squaring both sides}} \cr & \Rightarrow \frac{{3 + x}}{{3 - x}} = 9 \cr & \Rightarrow 3 + x = 27 - 9x \cr & \Rightarrow 10x = 24 \cr & \Rightarrow x = \frac{{24}}{{10}} \cr & \Rightarrow x = \frac{{12}}{5} \cr} $$

Solution:

 $$\eqalign{ & \root 3 \of a + \root 3 \of b = \root 3 \of c \cr & {\text{Take cube both sides}} \cr & {\left( {\root 3 \of a + \root 3 \of b } \right)^3} = {\left( {\root 3 \of c } \right)^3} \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {\root 3 \of a + \root 3 \of b } \right) = c \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} = c \cr & \Rightarrow a + b - c = - 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again take cube both sides}} \cr & \Rightarrow {\left( {a + b - c} \right)^3} = - 27abc \cr & \Rightarrow {\left( {a + b - c} \right)^3} + 27abc = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }} \cr & a = 0 \cr & b = 1 \cr & c = 1 \cr & {\text{Value of }}{\left( {a + b - c} \right)^3} + 27abc \cr & = {\left( {0 + 1 - 1} \right)^3} + 27 \times 0 \times 1 \times 1 \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 5 \cr & \left( {{\text{By squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = {\left( 5 \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = 23 \cr & {\text{Now,}} \cr & \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & {\text{Divided by }}{x^2}, \cr & \Rightarrow \frac{{\frac{{{x^4}}}{{{x^2}}} + \frac{{3{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + 3x + 5 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}} + 3\left( {x + \frac{1}{x}} \right) + 5}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{23 + 3\left( 5 \right) + 5}}{{23}} \cr & \Rightarrow \frac{{43}}{{23}} \cr} $$