Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$x + \frac{1}{x} = 1$$
x² + 1 - x = 0
(x + 1)(x² + 1² - x) = 0
x³ + 1 = 0
x³ = -1
⇒ x¹² + x⁹ + x⁶ + x³ + 1
= 1 - 1 + 1 - 1 + 1
= 1

(3 + 2√5)² = 29 + K√5
⇒ 9 + 20 + 12√5 = 29 + K√5
⇒ 29 + 12√5 = 29 + K√5
⇒ K√5 = 12√5
⇒ K = 12

$$\eqalign{ & \frac{{\left( {x + y + z} \right)\left( {xy + yz + zx} \right) - xyz}}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}} \cr & {\text{put }}z = 0, \cr & = \frac{{\left( {x + y + 0} \right)\left( {xy + 0 + 0} \right) - 0}}{{\left( {x + y} \right)\left( {y + 0} \right)\left( {0 + x} \right)}} \cr & = \frac{{\left( {x + y} \right)\left( {xy} \right)}}{{\left( {x + y} \right)\left( {xy} \right)}} \cr & = 1 \cr} $$

$$\eqalign{ & 3x\left( {x - \frac{5}{3} + \frac{1}{{3x}}} \right) = 0 \cr & x + \frac{1}{{3x}} = \frac{5}{3} \cr & {x^2} + \frac{1}{{9{x^2}}} + \frac{2}{3} = \frac{{25}}{9} \cr & {x^2} + \frac{1}{{9{x^2}}} = \frac{{25 - 6}}{9} \cr & {x^2} + \frac{1}{{9{x^2}}} = \frac{{19}}{9} \cr & {x^2} + \frac{1}{{9{x^2}}} = 2\frac{1}{9} \cr} $$

$$\eqalign{ & {\text{Given, }}{x^2} + \frac{1}{{{x^2}}} = \frac{7}{4} \cr & {\text{Squaring both sides, we get}} \cr & {x^4} + \frac{1}{{{x^4}}} + 2 = \frac{{49}}{{16}} \cr & {x^4} + \frac{1}{{{x^4}}} = \frac{{49}}{{16}} - 2 \cr & {x^4} + \frac{1}{{{x^4}}} = \frac{{49 - 32}}{{16}} \cr & {x^4} + \frac{1}{{{x^4}}} = \frac{{17}}{{16}} \cr} $$

$$\eqalign{ & {a^4} + 1 = \left( {\frac{{{a^2}}}{{{b^2}}}} \right)\left( {4{b^2} - {b^4} - 1} \right) \cr & {\text{take }}a = b = 1 \cr & 1 + 1 = \frac{1}{1}\left( {4 - 1 - 1} \right) \cr & 2 = 2{\text{ satisfied}} \cr & \therefore \,{a^4} + {b^4} = 1 + 1 = 2 \cr} $$

$$\eqalign{ & {x_1}{x_2}{x_3} = 4\left( {4 + {x_1} + {x_2} + {x_3}} \right) \cr & \left[ {\frac{1}{{2 + {x_1}}}} \right] + \left[ {\frac{1}{{2 + {x_2}}}} \right] + \left[ {\frac{1}{{2 + {x_3}}}} \right] = ? \cr & {\text{Assume value of }}{x_1},\,{x_2}\,\& \,{x_3} \cr & {x_1} = 4,\,{x_2} = 4,\,{x_3} = 4 \cr & 4 \times 4 \times 4 = 4\left( {4 + 4 + 4 + 4} \right) \cr & 64 = 64{\text{ value satisfied}} \cr & \therefore \left[ {\frac{1}{{2 + 4}} + \frac{1}{{2 + 4}} + \frac{1}{{2 + 4}}} \right] \cr & = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \cr & = \frac{3}{6} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & {a^3} + {b^3} + {c^3} = 3abc \cr & {\text{If }}a + b + c = 0 \cr & a = 0.324 \cr & b = 0.221 \cr & c = - 0.545 \cr & \frac{{\left( {0.545} \right)\left( {0.081} \right)\left( {0.51} \right)\left( {5.2} \right)}}{{3abc}} \cr & = - \frac{{0.545 \times 0.081 \times 0.51 \times 5.2}}{{3 \times 0.324 \times 0.221 \times 0.545}} \cr & = - \frac{{81 \times 510 \times 5.2}}{{3 \times 18 \times 18 \times 13 \times 17}} \cr & = - 1 \cr} $$

$$\eqalign{ & x + y + z = 19, \cr & xy + yz + zx = 144, \cr & \sqrt {{x^3} + {y^3} + {z^3} - 3xyz} \cr & {\text{Let }}z = 0 \cr & x + y = 19,\,xy = 144,\,\sqrt {{x^3} + {y^3}} = ? \cr & \sqrt {{x^3} + {y^3}} \cr & = \sqrt {\left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} - 3xy} \right]} \cr & = \sqrt {19\left( {{{19}^2} - 3 \times 144} \right)} \cr & = \sqrt {19 \times 19} \cr & = 19 \cr} $$

$$\eqalign{ & \sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1 \cr & {\text{Squaring both sides}} \cr & {a^2} + {b^2} + ab + {a^2} + {b^2} - ab + 2\sqrt {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} = 1 \cr & \sqrt {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} = \frac{1}{2} - \left( {{a^2} + {b^2}} \right) \cr & {\text{Again squaring both sides}} \cr & {a^4} + {b^4} + 2{a^2}{b^2} - {a^2}{b^2} = \frac{1}{4} + {\left( {{a^2} + {b^2}} \right)^2} - \left( {{a^2} + {b^2}} \right) \cr & {a^4} + {b^4} + {a^2}{b^2} = \frac{1}{4} + {a^4} + {b^4} + 2{a^2}{b^2} - {a^2} - {b^2} \cr & {a^2} + {b^2} - {a^2}{b^2} = \frac{1}{4}........\left( {\text{i}} \right) \cr & \Rightarrow \left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right) \cr & = 1 - {a^2} - {b^2} + {a^2}{b^2} \cr & = 1 - \left[ {{a^2} + {b^2} - {a^2}{b^2}} \right] \cr & = 1 - \frac{1}{4} \cr & = \frac{3}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}b = 0 \cr & \sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1 \cr & \sqrt {{a^2} + 0 + 0} + \sqrt {{a^2} + 0 + 0} = 1 \cr & a + a = 1 \cr & 2a = 1 \cr & a = \frac{1}{2} \cr & \left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right) \cr & = \left( {1 - \frac{1}{4}} \right)\left( {1 - 0} \right) \cr & = \frac{3}{4} \cr} $$