Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\text{ }}x = 3 + 2\sqrt 2 \cr & \Rightarrow {x^2} = {\left( {3 + 2\sqrt 2 } \right)^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} = 9 + 8 + 12\sqrt 2 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr & \Rightarrow \frac{1}{{{x^2}}} = \frac{1}{{17 + 12\sqrt 2 }} \times \frac{{17 - 12\sqrt 2 }}{{17 - 12\sqrt 2 }} \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 12\sqrt 2 \cr & \therefore {\text{ }}{x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 17 + 12\sqrt 2 + 17 - 12\sqrt 2 \cr & = 34 \cr} $$

$$\eqalign{ & x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x} \cr & \Rightarrow 3x - 2 = \frac{3}{x} \cr & \Rightarrow 3x - \frac{3}{x} = 2 \cr & \Rightarrow 3\left( {x - \frac{1}{x}} \right) = 2 \cr & \Rightarrow x - \frac{1}{x} = \frac{2}{3} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} - 2 = \frac{4}{9} \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = \frac{4}{9} + 2 \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = 2\frac{4}{9} \cr} $$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & \therefore {x^2} + x + \frac{1}{x} + \frac{1}{{{x^2}}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + x + \frac{1}{x} \cr & \Rightarrow 7 + 3 \cr & \Rightarrow 10 \cr} $$

$$\eqalign{ & {a^2} + {b^2} = 5ab \cr & \Rightarrow \frac{{{a^2}}}{{ab}} + \frac{{{b^2}}}{{ab}} = 5 \cr & \Rightarrow \frac{a}{b}{\text{ + }}\frac{b}{a}{\text{ = 5}} \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^2}{\text{ + }}{\left( {\frac{b}{a}} \right)^2} + 2 \times \frac{a}{b} \times \frac{b}{a} = 25 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 25 - 2 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 23 \cr} $$

$$\eqalign{ & xy + yz + zx = 0 \cr & \therefore xy + zx = - yz \cr & \Rightarrow xy + yz = - zx \cr & \Rightarrow yz + zx = - xy \cr & \therefore \frac{1}{{{x^2} - yz}} + \frac{1}{{{y^2} - zx}} + \frac{1}{{{z^2} - xy}} \cr} $$
Putting values of -yz, -zx, -xy from above
$$ \Rightarrow \frac{1}{{{x^2} + \left( {xy + zx} \right)}} + \frac{1}{{{y^2} + \left( {xy + yz} \right)}}$$       $$ + \frac{1}{{{z^2} + \left( {yz + zx} \right)}}$$
$$ \Rightarrow \frac{1}{{x\left( {x + y + z} \right)}} + \frac{1}{{y\left( {x + y + z} \right)}}$$       $$ + \frac{1}{{z\left( {x + y + z} \right)}}$$
$$\eqalign{ & \Rightarrow \frac{1}{{\left( {x + y + z} \right)}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) \cr & \Rightarrow \frac{1}{{\left( {x + y + z} \right)}}\left( {\frac{{zy + xz + xy}}{{xyz}}} \right) \cr & \Rightarrow \frac{1}{{x + y + z}} \times 0 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & a + b + c = 9 \cr & {\text{For minimum value}} \cr & a = b = c \cr & \Rightarrow 3a = 9 \cr & \Rightarrow a = \frac{9}{3} \cr & \Rightarrow a = 3 \cr & {\text{For minimum value}} \cr & a = b = c = 3 \cr & \therefore {a^2} + {b^2} + {c^2} \cr & \Rightarrow {3^2} + {3^2} + {3^2} \cr & \Rightarrow 9 + 9 + 9 \cr & \Rightarrow 27 \cr} $$

$$\eqalign{ & {a^2} + {b^2} + 4{c^2} = 2\left( {a + b - 2c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + 4{c^2} - 2a - 2b + 4c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} - 2b + 1 + 4{c^2} + 4c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {2c + 1} \right)^2} = 0 \cr & \cr & \therefore a - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 1 \cr & \,\,\,\,\,\,\,b - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = 1 \cr & \,\,\,\,2c + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = \frac{{ - 1}}{2} \cr & \cr & \therefore {a^2} + {b^2} + {c^2} \cr & \Rightarrow 1 + 1 + \frac{1}{4} \cr & \Rightarrow 2 + \frac{1}{4} \cr & \Rightarrow \frac{9}{4} \cr & \Rightarrow 2\frac{1}{4} \cr} $$

$$\frac{{x - {a^2}}}{{b + c}} + \frac{{x - {b^2}}}{{c + a}} + \frac{{x - {c^2}}}{{a + b}} = 4\left( {a + b + c} \right)$$
Note : In such type of question to save your valuable time assume values as per your need which make your calculation easier.
Assume a = 1, b = 0, c = 1
$$\eqalign{ & {\text{Make sure there will be no }}\left( {\frac{0}{0}} \right){\text{ form}} \cr & \therefore \frac{{x - 1}}{{1 + 0}} + \frac{{x - 0}}{{1 + 1}} + \frac{{x - 1}}{{1 + 0}} = 4 \cr & \Rightarrow x - 1 + \frac{x}{2} + x - 1 = 4 \times 2 \cr & \Rightarrow x + \frac{x}{2} + x = 8 + 2 \cr & \Rightarrow \frac{{5x}}{2} = 10 \cr & \Rightarrow x = 4 \cr & {\text{Now put values in options take option}} \cr & \left( \text{A} \right),{\left( {a + b + c} \right)^2} = {\left( {1 + 0 + 1} \right)^2} = 4 \cr} $$

$$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{4}{5}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{{15}}{{16}} \cr & \Rightarrow \frac{a}{b} \times \frac{b}{c}{\text{ = }}\frac{4}{5} \times \frac{{15}}{{16}} \cr & \Rightarrow \frac{a}{c} = \frac{3}{4} \cr & \therefore \frac{{18{c^2} - 7{a^2}}}{{45{c^2} + 20{a^2}}} \cr & = \frac{{{c^2}\left( {18 - 7\frac{{{a^2}}}{{{c^2}}}} \right)}}{{{c^2}\left( {45 + 20\frac{{{a^2}}}{{{c^2}}}} \right)}} \cr & = \frac{{18 - 7{{\left( {\frac{a}{c}} \right)}^2}}}{{45 + 20{{\left( {\frac{a}{c}} \right)}^2}}} \cr & = \frac{{18 - 7 \times \frac{9}{{16}}}}{{45 + 20 \times \frac{9}{{16}}}} \cr & = \frac{{18 - \frac{{63}}{{16}}}}{{45 + \frac{{45}}{4}}} \cr & = \frac{{225 \times 4}}{{16 \times 225}} \cr & = \frac{1}{4} \cr} $$

$$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}}$$
Go through option D take x = y = z
$$\eqalign{ & \frac{1}{{{x^2}}} + \frac{1}{{{x^2}}} + \frac{1}{{{x^2}}} = \frac{1}{{x^2}} + \frac{1}{{x^2}} + \frac{1}{{x^2}} \cr & \therefore {\text{Option 'D' is right}} \cr} $$