Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

x⁴ + x² + 25
Check by the option,
(x² - 3x + 5)(x² - 3x + 5)
= x⁴ - 3x³ + 5x² + 3x³ - 9x² + 15x + 5x² - 15x + 25
= x⁴ + x² + 25

Other Method
Let the value of x = 1
then, x⁴ + x² + 25 = 27
Now check options, and there is only one option that is equal to 27.
(1 + 3 + 5)(1 - 3 + 5) = 9 × 3 = 27

[(a² - b²)³ + (b² - c²)³ + (c² - a²)³] ÷ [(a - b)³ + (b - c)³ + (c - a)³]
put a = 0, b = 1, c = 2
$$\frac{{ - 1 - 27 + 64}}{{ - 1 - 1 + 8}} = \frac{{36}}{6} = 6$$
(0 + 1)(1 + 2)(2 + 0) = 3 × 2 = 6
Hence option A is right answer.

$$\eqalign{ & \frac{{\left( {{a^2} + {b^2}} \right)\left( {a - b} \right) - \left( {{a^3} - {b^3}} \right)}}{{{a^2}b - a{b^2}}} \cr & = \frac{{\left( {{a^2} + {b^2}} \right)\left( {a - b} \right) - \left( {{a^3} - {b^3}} \right)}}{{ab\left( {a - b} \right)}} \cr & = \frac{{\left( {a - b} \right)\left( {{a^2} + {b^2} - {a^2} - {b^2} - ab} \right)}}{{ab\left( {a - b} \right)}} \cr & = - 1 \cr} $$

$$\eqalign{ & {x^4} + \frac{1}{{{x^4}}} = 14159 \cr & {x^2} + \frac{1}{{{x^2}}} = \sqrt {14159 + 2} \cr & {x^2} + \frac{1}{{{x^2}}} = 119 \cr & x + \frac{1}{x} = \sqrt {119 + 2} \cr & x + \frac{1}{x} = 11 \cr} $$

$$\eqalign{ & 2x + \frac{1}{{2x}} = 2 \cr & {\text{Now, }}2x = 1 \cr & \Rightarrow x = \frac{1}{2} \cr & \Rightarrow \frac{1}{x} = 2 \cr & \sqrt {2{{\left( {\frac{1}{x}} \right)}^4} + {{\left( {\frac{1}{x}} \right)}^5}} \cr & = \sqrt {2 \times {2^4} + {2^5}} \cr & = \sqrt {32 + 32} \cr & = \sqrt {64} \cr & = 8 \cr} $$

$$\eqalign{ & x = 32.5,\,y = 34.6{\text{ and }}z = 30.9 \cr & {x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left[ {\frac{1}{2}\left\{ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right\}} \right] \cr & = \left( {32.5 + 34.6 + 30.9} \right)\left[ {\frac{1}{2}\left\{ {{{\left( {32.5 - 34.6} \right)}^2} + {{\left( {34.6 - 30.9} \right)}^2} + {{\left( {30.9 - 32.5} \right)}^2}} \right\}} \right] \cr & = 98\left[ {\frac{1}{2}\left\{ {{{\left( { - 2.1} \right)}^2} + {{\left( {3.7} \right)}^2} + {{\left( { - 1.6} \right)}^2}} \right\}} \right] \cr & = 98\left[ {\frac{1}{2}\left\{ {4.41 + 13.69 + 2.56} \right\}} \right] \cr & = 98\left[ {\frac{1}{2}\left\{ {20.66} \right\}} \right] \cr & = 98 \times 10.33 \cr & = 1012.34 \cr & {x^3} + {y^3} + {z^3} - 3xyz = 0.98k \cr & 1012.34 = 0.98k \cr & k = \frac{{1012.34}}{{0.98}} = 1033 \cr} $$

a + b = 5c
$$\frac{{{a^2} + {b^2} + {c^2} - 1024}}{{ab - bc - ca}} = - 2$$
a² + b² + c² + 2ab - 2bc - 2ac = 1024
(a + b)² + c(c - 2b - 2a) = 1024
(5c)² + c[c - 2(b + a)] = 1024
25c² + c[c - 2 × 5c) = 1024
25c² - 9c² = 1024
16c² = 1024
c² = 64
c = 8

$$\eqalign{ & x = 2 + \sqrt 3 \cr & y = 2 - \sqrt 3 \cr & z = 1 \cr & \frac{x}{{yz}} + \frac{y}{{xz}} + \frac{z}{{xy}} + 2\left[ {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right] \cr & = \frac{{{x^2} + {y^2} + {z^2}}}{{xyz}} + 2\left[ {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right] \cr & = \frac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2}}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\left( 1 \right)}} + 2\left[ {\frac{1}{{\left( {2 + \sqrt 3 } \right)}} + \frac{1}{{\left( {2 - \sqrt 3 } \right)}} + 1} \right] \cr & = \frac{{4 + 3 + 4\sqrt 3 + 4 + 3 - 4\sqrt 3 + 1}}{1} + 2\left[ {2 - \sqrt 3 + 2 + \sqrt 3 + 1} \right] \cr & = 15 + 10 \cr & = 25 \cr} $$

$$\eqalign{ & \frac{1}{{{x^2} + {a^2}}} = {x^2} - {a^2} \cr & 1 = {x^4} - {a^4} \cr & {x^4} = 1 + {a^4} \cr & x = {\left( {1 + {a^4}} \right)^{\frac{1}{4}}} \cr} $$

xy = 2
2 × 1
1 × 2
at x = 2, y = 1
9x² + y² = 37
9 × 4 + 1 = 37
37 = 37 it satisfy
27x³ + y³
= 27 × 2³ + 1³
= 27 × 8 + 1
= 217