Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & x + y + z = 0 \cr & {\text{Let }}x = y = 1 \cr & z = - 2 \cr & \frac{{{x^2} + {y^2} + {z^2}}}{{{z^2} - xy}} \cr & = \frac{{1 + 1 + 4}}{{4 - 1}} \cr & = 2 \cr} $$

$$\eqalign{ & \frac{{0.325 \times 0.325 + 0.175 \times 0.175 + 25 \times 0.00455}}{{5 \times 0.0065 \times 3.25 - 7 \times 0.175 \times 0.025}} - \frac{{0.5}}{{1.5}} \cr & = \frac{{{{\left( {3.35 + 0.175} \right)}^2}}}{{\left( {0.325 - 0.175} \right)\left( {0.325 + 0.175} \right)}} - \frac{1}{3} \cr & = \frac{{0.500}}{{0.150}} - \frac{1}{3} \cr & = \frac{{10}}{3} - \frac{1}{3} \cr & = \frac{9}{3} \cr & = 3 \cr} $$

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

$$\eqalign{ & {x^4} - 83{x^2} + 1 = 0 \cr & {x^2} - 83 + \frac{1}{{{x^2}}} = 0 \cr & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & x - \frac{1}{x} = 9 \cr & {\left( {x - \frac{1}{x}} \right)^3} = {9^3} \cr & {x^3} - \frac{1}{{{x^3}}} - 3 \times 9 = 729 \cr & {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$

1 + 9r² + 81r⁴ = 256
1 + 3r + 9r² = 32
So, 1 - 3r + 9r² $$ = \frac{{256}}{{32}} = 8$$

$$\eqalign{ & {\text{Given, }}{x^2} + \frac{1}{{{x^2}}} = \frac{7}{4} \cr & {\text{Adding 2 to both sides}} \cr & {x^2} + \frac{1}{{{x^2}}} + 2 = \frac{7}{4} + 2 \cr & {\left( {x + \frac{1}{x}} \right)^2} = \frac{{15}}{4} \cr & x + \frac{1}{x} = \frac{{\sqrt {15} }}{2} \cr & \because \,{x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) \cr & \therefore \,{x^3} + \frac{1}{{{x^3}}} = \frac{{15 \times \sqrt {15} }}{8} - 3 \times \frac{{\sqrt {15} }}{2} \cr & = \frac{{15\sqrt {15} - 12\sqrt {15} }}{8} \cr & = \frac{{3\sqrt {15} }}{8} \cr} $$

$$\eqalign{ & {y^4} + \frac{1}{{{y^4}}} = 223 \cr & \Rightarrow {y^2} + \frac{1}{{{y^2}}} = \sqrt {223 + 2} \cr & \Rightarrow {y^2} + \frac{1}{{{y^2}}} = \sqrt {225} \cr & \Rightarrow {y^2} + \frac{1}{{{y^2}}} = 15 \cr} $$

$$\eqalign{ & \frac{{{y^2}}}{{2xz}} - \frac{{{x^2}}}{{2yz}} - \frac{{{z^2}}}{{2xy}} \cr & \frac{{{y^3} - {x^3} - {z^3}}}{{2xyz}}\,......\,\left( 1 \right) \cr & x - y + z = 0 \cr & x + z = y \cr & {\text{Cubing both side}} \cr & {\left( {x + z} \right)^3} = {y^3} \cr & {x^3} + {z^3} + 3\left( {x + z} \right)\left( x \right)\left( z \right) = {y^3} \cr & {x^3} + {z^3} + 3\left( y \right)\left( x \right)\left( z \right) = {y^3} \cr & 3xyz = {y^3} - {x^3} - {z^3} \cr & {\text{Put in equation }}\left( 1 \right) \cr & \frac{{3xyz}}{{2xyz}} = \boxed{\frac{3}{2}} \cr} $$

$$\eqalign{ & {x^2} - 5x + 1 = 0 \cr & \left( {{x^4} + \frac{1}{{{x^2}}}} \right) \div \left( {{x^2} + 1} \right) = ? \cr & {x^2} - 5x + 1 = 0 \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 110 \cr & \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{x\left( {x + \frac{1}{x}} \right)}} = \frac{{110}}{5} = 22 \cr} $$

Let c = 0
a + b = 6
a³ + b³ = 342
ab = ?
(a³ + b³) = (a + b)[(a + b)² - 3ab]
342 = 6[6² - 3ab]
57 = 36 - 3ab
3ab = -21
ab = -7