Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

2x² + y² + 6x - 2xy + 9 = 0
x² + x² + y² + 6x - 2xy + 9= 0
(x² + y² - 2xy) + (x² + 6x + 9) = 0
(x - y)² + (x + 3)² = 0
x = y, x = -3
4x³ - y³ + x²y²
= 4x³ - x³ + x⁴
= 3x³ + x⁴
= -x⁴ + x⁴
= 0

$$\eqalign{ & \frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3 + \sqrt 5 } }} = a + \sqrt 5 b \cr & {\text{LHS}} = \frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3 + \sqrt 5 } }} \cr & = \frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {\frac{{6 + 2\sqrt 5 }}{2}} }} \cr & = \frac{{22\sqrt 2 }}{{4\sqrt 2 - \frac{{\sqrt 5 + 1}}{{\sqrt 2 }}}} \cr & = \frac{{22\sqrt 2 }}{{\frac{{8 - \sqrt 5 - 1}}{{\sqrt 2 }}}} \cr & = \frac{{22 \times 2}}{{7 - \sqrt 5 }} \times \frac{{7 + \sqrt 5 }}{{7 + \sqrt 5 }} \cr & = \frac{{44\left( {7 + \sqrt 5 } \right)}}{{{7^2} - {{\left( {\sqrt 5 } \right)}^2}}} \cr & = \frac{{44\left( {7 + \sqrt 5 } \right)}}{{44}} \cr & = 7 + \sqrt 5 \cr & {\text{Compare LHS and RHS}} \cr & 7 + \sqrt 5 = a + \sqrt 5 b \cr & a = 7;\,b = 1 \cr & \left( {ab} \right):\left( {a + b} \right) = \left( {7 \times 1} \right):\left( {7 + 1} \right) = 7:8 \cr} $$

$$\eqalign{ & 2x - y = 2{\text{ and }}xy = \frac{3}{2} \cr & 2x - y = 2 \cr & {\text{divide '2' both sides}} \cr & x - \frac{y}{2} = 1 \cr & {\text{cube both sides}} \cr & {\left( {x - \frac{y}{2}} \right)^3} = {1^3} \cr & {x^3} - \frac{{{y^3}}}{8} - 3 \times x \times \frac{y}{2}\left( {x - \frac{y}{2}} \right) = {1^3} \cr & {x^3} - \frac{{{y^3}}}{8} = {1^3} + 3 \times \frac{3}{2} \times \frac{1}{2} \times 1 \cr & {x^3} - \frac{{{y^3}}}{8} = 1 + \frac{9}{4} \cr & {x^3} - \frac{{{y^3}}}{8} = \frac{{13}}{4} \cr} $$

$$\eqalign{ & \frac{{4\left( {\frac{{2x}}{5} - \frac{3}{2}} \right)}}{3} + \frac{7}{5} = \frac{{37}}{5} \cr & \Rightarrow \frac{{\frac{{8x}}{5} - 6}}{3} + \frac{7}{5} = \frac{{37}}{5} \cr & \Rightarrow 8x - 30 + 21 = 111 \cr & \Rightarrow 8x = 111 + 9 \cr & \Rightarrow x = \frac{{120}}{8} \cr & \Rightarrow x = 15 \cr} $$

$$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \frac{1}{x} = 3 - 2\sqrt 2 \cr & x + \frac{1}{x} = 6 \cr & {x^2} + \frac{1}{{{x^2}}} = {\left( 6 \right)^2} - 2 \cr & {x^2} + \frac{1}{{{x^2}}} = 34 \cr} $$

$$\eqalign{ & p{\left( {x + y} \right)^2} = 5{\text{ and }}q{\left( {x - y} \right)^2} = 3 \cr & {\text{Put the value of }}x = 2{\text{ and }}y = 1 \cr & p{\left( {2 + 1} \right)^2} = 5{\text{ and }}q{\left( {2 - 1} \right)^2} = 3 \cr & p = \frac{5}{9},\,q = 3 \cr & \to {p^2}{\left( {x + y} \right)^2} + 4pqxy - {q^2}{\left( {x - y} \right)^2} \cr & = {\left( {\frac{5}{9}} \right)^2}{\left( {2 + 1} \right)^2} + 4 \times \frac{5}{9} \times 3 \times 2 \times 1 - {\left( 3 \right)^2}{\left( {2 - 1} \right)^2} \cr & = \frac{{25}}{{81}} \times 9 + \frac{{40}}{3} - 9 \cr & = \frac{{25}}{9} + \frac{{40}}{3} - 9 \cr & = \frac{{25 + 120 - 81}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Put the value of }}p{\text{ and }}q{\text{ in option A}} \cr & {\text{option A }} \to {\text{2}}\left( {p + q} \right) \cr & = 2\left( {\frac{5}{9} + 3} \right) \cr & = 2 \times \frac{{32}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Option A is satisfied}} \cr & {\text{So, 2}}\left( {p + q} \right)\,{\text{is answer}} \cr} $$

x² - 3x + 2 = 0
⇒ x² - 2x - x + 2 = 0
⇒ x(x - 2) - 1(x - 2) = 0
⇒ (x - 2)(x - 1) = 0
x = 2 and 1
When x = 2, then
x³ - 6x² + ax + b = 0
⇒ 8 - 24 + 2a + b = 0
⇒ 2a + b = 16 . . . . . . (i)
When x = 1, then
x³ - 6x² + ax + b = 0
⇒ 1 - 6 + a + b = 0
⇒ a + b = 5 . . . . . . (ii)
From equation (ii)
a + 5 = 16
a = 11, b = -6

$$\eqalign{ & {x^8} - 1442{x^4} + 1 = 0 \cr & \frac{{{x^8}}}{{{x^4}}} - \frac{{1442{x^4}}}{{{x^4}}} + \frac{1}{{{x^4}}} = 0 \cr & {x^4} - 1442 + \frac{1}{{{x^4}}} = 0 \cr & {x^4} + \frac{1}{{{x^4}}} = 1442 \cr & {x^4} + \frac{1}{{{x^4}}} + 2 = 1444 \cr & \left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 38 \cr & {x^2} + \frac{1}{{{x^2}}} - 2 = 36 \cr & {\left( {x - \frac{1}{x}} \right)^2} = 36 \cr & x - \frac{1}{x} = 6 \cr} $$

1006² - 1007 × 1005 + 1008 × 1004 - 1009 × 1003
= 1006² - (1006 + 1)(1006 -1) + (1006 + 2)(1006 - 2) - (1006 + 3)(1006 - 3)
= 1006² - 1006² + 1 + 1006² - 4 - 1006² + 9
= 6

$$\eqalign{ & \frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}} \cr & = \frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \times \frac{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}} \cr & = \frac{{203 \times 292}}{{7 \times 365 \times 29}} \times \frac{{0.25\left( {0.25 + 19.99} \right)}}{{\left( {20.24 \times 4} \right)}} \cr & = \frac{{29 \times 292}}{{365 \times 29}} \times \frac{{0.25 \times 2024}}{{20.24 \times 4}} \cr & = \frac{{292}}{{365}} \times \frac{1}{{16}} \cr & = 0.05 \cr} $$