Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \left( {8{x^3} - 27{y^3}} \right) \div \left( {2x - 3y} \right) = A{x^2} + Bxy + C{y^2} \cr & \frac{{\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)}}{{\left( {2x - 3y} \right)}} = A{x^2} + Bxy + C{y^2} \cr & 4{x^2} + 6xy + 9{y^2} = A{x^2} + Bxy + C{y^2} \cr & {\text{Comparison both side}} \cr & A = 4,\,B = 6,\,C = 9 \cr & \left( {2A + B - C} \right) \cr & = \left( {2 \times 4 + 6 - 9} \right) \cr & = 5 \cr} $$

$$\eqalign{ & k\left( {21{x^2} + 24} \right) + rx + \left( {14{x^2} - 9} \right) = 0\,......\left( {\text{i}} \right) \cr & k\left( {7{x^2} + 8} \right) + px + \left( {2{x^2} - 3} \right) = 0\,......\left( {{\text{ii}}} \right) \cr & {\text{from }}\left( {\text{i}} \right), \cr & 21k{x^2} + 24k + rx + 14{x^2} - 9 = 0 \cr & \left( {21k + 14} \right){x^2} + rx + \left( {24k - 9} \right) = 0\,......\left( {{\text{iii}}} \right) \cr & {\text{from }}\left( {{\text{ii}}} \right), \cr & 7k{x^2} + 8k + px + 2{x^2} - 3 = 0 \cr & \left( {7k + 2} \right){x^2} + px + \left( {8k - 3} \right) = 0\,......\left( {{\text{iv}}} \right) \cr & {\text{If roots are common then,}} \cr & \frac{{21k + 14}}{{7k + 2}} = \frac{r}{p} = \frac{{24k - 9}}{{8k - 3}} \cr & \Rightarrow \frac{r}{p} = \frac{{3\left( {8k - 3} \right)}}{{\left( {8k - 3} \right)}} \cr & \Rightarrow \frac{r}{p} = \frac{3}{1} \cr & \Rightarrow \frac{p}{r} = \frac{1}{3} \cr} $$

$$\eqalign{ & xy = - 6,\,{x^3} + {y^3} = 19 \cr & {\text{putting }}x = 3,\,y = - 2 \cr & {\text{both condition are satisfied}} \cr & {\text{So, }}\frac{1}{{{x^{ - 1}}}} + \frac{1}{{{y^{ - 1}}}} \cr & = x + y \cr & = 3 - 2 \cr & = 1 \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} + 84 = 4\left( {a - 2b + 4c} \right) \cr & a = 2 \cr & b = - 4 \cr & c = 8 \cr & = \sqrt {ab - bc + ca} \cr & = \sqrt {2 \times \left( { - 4} \right) - \left( { - 4} \right) \times 8 + 8 \times 2} \cr & = \sqrt {32 + 8} \cr & = \sqrt {40} \cr & = 2\sqrt {10} \cr} $$

$$\eqalign{ & 4.6 = b \cr & 5.4 = a \cr & \frac{{{b^4} + {a^4} + {a^2}{b^2}}}{{{a^2} + {b^2} + ab}} \cr & = \frac{{\left( {{a^2} + {b^2} + ab} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{{a^2} + {b^2} + ab}} \cr & = {a^2} + {b^2} - ab \cr & = {\left( {a - b} \right)^2} + ab \cr & = {\left( {5.4 - 4.6} \right)^2} + 24.84 \cr & = {0.8^2} + 24.84 \cr & = 25.48 \cr} $$

ab + bc + ca = 8
a² + b² + c² = 20
a² + b² + c² - ab - bc - ca = $$\frac{1}{2}$$[(a - b)² + (b - c)² + (c - a)²]
20 - 8 = $$\frac{1}{2}$$[(a - b)² + (b - c)² + (c - a)²)]
12 = $$\frac{1}{2}$$[(a - b)² + (b - c)² + (c - a)²]
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
a + b + c = $$\sqrt {20 + 2 \times 8} = 6$$
$$\frac{1}{2}$$(a + b + c)[(a - b)² + (b - c)² + (c - a)²]
= 12(a + b+ c)
= 12 × 6
= 72

$$\eqalign{ & {a^2} + {b^2} + {c^2} + 27 = 6\left( {a + b + c} \right) \cr & {\left( {a - 3} \right)^2} + {\left( {b - 3} \right)^2} + {\left( {c - 3} \right)^2} = 0 \cr & a = 3,\,b = 3,\,c = 3 \cr & \root 3 \of {{a^3} + {b^3} - {c^3}} \cr & = \root 3 \of {27 + 27 - 27} \cr & = \root 3 \of {27} \cr & = 3 \cr} $$

$$\eqalign{ & {x^2} - \sqrt 7 x + 1 = 0 \cr & {x^2} + 1 = \sqrt 7 x \cr & x + \frac{1}{x} = \sqrt 7 \cr & {x^5} + \frac{1}{{{x^5}}} = \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) - \left( {x + \frac{1}{x}} \right) \cr & = \left( {{{\left( {\sqrt 7 } \right)}^2} - 2} \right)\left( {{{\left( {\sqrt 7 } \right)}^3} - 3 \times \sqrt 7 } \right) - \sqrt 7 \cr & = 5 \times 4\sqrt 7 - \sqrt 7 \cr & = 19\sqrt 7 \cr} $$

$$\eqalign{ & x + \frac{1}{{16x}} = 3 \cr & 2x + \frac{1}{{8x}} = 6 \cr & {\text{Cube both side}} \cr & 8{x^3} + \frac{1}{{512{x^3}}} + 3 \times 2 \times \frac{1}{8} \times 6 = 216 \cr & 8{x^3} + \frac{1}{{512{x^3}}} = 216 - \frac{9}{2} \cr & {\text{Multiply by '2' both side}} \cr & 16{x^3} + \frac{1}{{256{x^3}}} = 432 - 9 = 423 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 5 \cr & {\text{Cubing both sides, we get}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 125 \cr & {x^3} + \frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & {x^3} + \frac{1}{{{x^3}}} = 110 \cr & {\text{Squaring both sides, we get}} \cr & {x^6} + \frac{1}{{{x^6}}} + 2 = 12100 \cr & {x^6} + \frac{1}{{{x^6}}} = 12098 \cr} $$