Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 66 \cr & \text{Subtract 2 from both sides} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 66 - 2 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 64 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = {\left( 8 \right)^2} \cr & \Rightarrow x - \frac{1}{x} = \pm 8 \cr & \Rightarrow \frac{{{x^2} - 1 + 2x}}{x} \cr & \Rightarrow \frac{{\frac{{{x^2}}}{x} - \frac{1}{x} + \frac{{2x}}{x}}}{{\frac{x}{x}}} \cr & \Rightarrow \frac{{\left( {x - \frac{1}{x}} \right) + 2}}{1} \cr & {\text{When , }}x - \frac{1}{x} = + 8 \cr & {\text{Then,}} \cr & \left( {x - \frac{1}{x}} \right) + 2 = 8 + 2 = 10 \cr & {\text{When }}x - \frac{1}{x} = - 8 \cr & \Rightarrow x - \frac{1}{x} = - 8 + 2 \cr & \Rightarrow x - \frac{1}{x} = - 6 \cr & \therefore \left( {10, - 6} \right) \cr} $$

$${a^2} + a + 1 = 0$$
\[\left[ \begin{array}{l} {a^3} + {1^3} = \left( {a + 1} \right)\left( {{a^2} + a + 1} \right)\\ {a^3} - {1^3} = \left( {a - 1} \right)\left( {{a^2} + a + 1} \right) \end{array} \right]\]
$$\eqalign{ & \therefore \left( {{a^3} - 1} \right) = \left( {a - 1} \right) \times 0 \cr & \Rightarrow {a^3} - 1 = 0 \cr & \Rightarrow {a^3} = 1 \cr & \Rightarrow {\left( {{a^3}} \right)^3} = {1^3} \cr & \Rightarrow {a^9} = 1 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & x = - 2k{\text{ and }}y = 1 - 3k \cr & \therefore {\text{For }}x = y \cr & \Rightarrow - 2k = 1 - 3k \cr & \Rightarrow k = 1 \cr} $$

$$\eqalign{ & {\text{ }}x + \frac{1}{x} = 5 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( 5 \right)^3} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 110 \cr & \therefore {\text{Squaring both sides}} \cr & \Rightarrow {\left( {{x^3}{\text{ + }}\frac{1}{{{x^3}}}} \right)^2} = {\left( {110} \right)^2} \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} + 2 = 12100 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12100 - 2 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12098 \cr} $$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}{\text{ }} \cr & = \frac{{{x^6}}}{{{x^3}}} + \frac{{{x^4}}}{{{x^3}}} + \frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}} \cr & = {x^3} + \frac{1}{{{x^3}}} + \frac{1}{x} + x \cr & = 18 + 3 \cr & = 21 \cr} $$

$$\eqalign{ & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 \cr & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0{\text{ }} \cr & \Rightarrow \frac{p}{a} = x,{\text{ }}\frac{q}{b} = y,{\text{ }}\frac{r}{c} = z \cr & \Rightarrow \left( {x + y + z} \right) = 1 \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 1 \cr & {\text{and }}\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \cr & \Rightarrow \frac{{xy + yz + zx}}{{xyz}} = 0 \cr & \Rightarrow xy + yz + zx = 0 \cr & \therefore {x^2} + {y^2} + {z^2} = 1 \cr & {\text{So, }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1 \cr} $$

$$\eqalign{ & \frac{{{{\left( {x + 1} \right)}^3} - {{\left( {x - 1} \right)}^3}}}{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}} = 2 \cr & {{\text{A}}^3} - {{\text{B}}^3} = \left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr & {{\text{A}}^2} - {{\text{B}}^2} = \left( {{\text{A}} - {\text{B}}} \right)\left( {{\text{A}} + {\text{B}}} \right) \cr} $$
$$ \Rightarrow \frac{{\left( {x + 1 - x + 1} \right)\left\{ {{{\left( {x + 1} \right)}^2} + \left( {x - 1} \right)\left( {x + 1} \right) + {{\left( {x - 1} \right)}^2}} \right\}}}{{\left( {x + 1 - x + 1} \right)\left( {x + 1 + x - 1} \right)}} = 2$$

$$\eqalign{ & \Rightarrow \frac{{\left( {{x^2} + 1 + 2x + {x^2} - 1 + {x^2} + 1 - 2x} \right)}}{{\left( {2x} \right)}} = 2 \cr & \Rightarrow \frac{{3{x^2} + 1}}{{2x}} = 2 \cr & \Rightarrow 3{x^2} + 1 = 4x \cr & \Rightarrow 3{x^2} - 4x + 1 = 0 \cr & \Rightarrow 3{x^2} - 3x - x + 1 = 0 \cr & \Rightarrow 3x\left( {x - 1} \right) - 1\left( {x - 1} \right) = 0 \cr & \Rightarrow \left( {3x - 1} \right)\left( {x - 1} \right) = 0 \cr & 3x - 1 = 0 \cr & \Leftrightarrow x = \frac{1}{3} \cr & x - 1 = 0 \cr & \text{for } x = 1 = \frac{1}{1} \cr & {\text{By adding numinator and denominator }} \cr & 1 + 1 = 2 \cr & {\text{No option is satisfied}} \cr & \therefore x = \frac{1}{3} \cr & \Rightarrow x = 1 + 3 \cr & \Rightarrow x = 4 \cr} $$

$$x = \sqrt 5 + 2$$
Rationalization of denominator.
$$\eqalign{ & \Rightarrow \frac{1}{x} = \frac{1}{{\sqrt 5 + 2}} \times \frac{{\sqrt 5 - 2}}{{\sqrt 5 - 2}} \cr & \Rightarrow \frac{1}{x} = \frac{{\sqrt 5 - 2}}{{5 - 4}} \cr & \Rightarrow \frac{1}{x} = \sqrt 5 - 2 \cr & \Rightarrow x - \frac{1}{x} = \sqrt 5 + 2 - \sqrt 5 + 2 \cr & \Rightarrow x - \frac{1}{x} = 4 \cr & \therefore \frac{{2{x^2} - 3x - 2}}{{3{x^2} - 4x - 3}} \cr} $$
Multiply by $$\frac{1}{x}$$ in numerator and denominator
$$\eqalign{ & = \frac{{\frac{{2{x^2}}}{x} - \frac{{3x}}{x} - \frac{2}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} - \frac{3}{x}}} \cr & = \frac{{2x - \frac{2}{x} - 3}}{{3x - \frac{3}{x} - 4}} \cr & = \frac{{2\left( {x - \frac{1}{x}} \right) - 3}}{{3\left( {x - \frac{1}{x}} \right) - 4}} \cr & = \frac{{2 \times 4 - 3}}{{3 \times 4 - 4}} \cr & = \frac{{8 - 3}}{{12 - 4}} \cr & = \frac{5}{8} \cr & = 0.625 \cr} $$

$$\eqalign{ & {x^2} + {y^2} + 1 = 2x \cr & {x^2} - 2x + 1 + {y^2} = 0 \cr & {\left( {x - 1} \right)^2} + {y^2} = 0 \cr & {\text{If }}{{\text{A}}^2} + {{\text{B}}^2} = 0 \cr} $$
[As powers are even it can possible only when A = 0 & B = 0]
$$\eqalign{ & \therefore x - 1 = 0 \cr & x = 1 \cr & y = 0 \cr & \therefore {x^3} + {y^5} \cr & = 1 + 0 \cr & = 1 \cr} $$

$$\eqalign{ & x\left( {x - 3} \right) = - 1 \cr & \Rightarrow \left( {x - 3} \right) = \frac{{ - 1}}{x} \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - 3} \right)^3} = {\left( {\frac{{ - 1}}{x}} \right)^3} \cr & \Rightarrow {x^3} - 27 - 9.x.\left( {x - 3} \right) = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 - 9 \times - 1 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 + 9 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 18 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3}\left( {{x^3} - 18} \right) = - 1 \cr} $$