Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

x + y + z = 19, x² + y² + z² = 133 and xz = y²
(x - z) = ?
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
361 = 133 + 2(xy + yz + y²)
228 = 2(x + y + z)y
$$y = \frac{{114}}{{19}} = 6$$
x + z = 13
xz = 36
x - z = ?
(x + z)² - (x - z)² = 4xz
169 - (x - z)² = 144
x - z = 5

$$\eqalign{ & \because \,c - d = \frac{{c + d}}{5} = \frac{{cd}}{3} \cr & \Rightarrow 5c - 5d = c + d \cr & \Rightarrow 4c = 6d \cr & \Rightarrow \boxed{\frac{c}{d} = \frac{3}{2}} \cr & \Rightarrow c = \frac{3}{2}d\,......\,\left( 1 \right) \cr & \because \,\frac{{c + d}}{5} = \frac{{cd}}{3} \cr & \Rightarrow \frac{{\frac{3}{2}d + d}}{5} = \frac{{\frac{3}{2}{d^2}}}{3} \cr & \Rightarrow \frac{{5d}}{2} \times \frac{1}{5} = \frac{1}{2}{d^2} \cr & \Rightarrow d = 1 \cr & {\text{From equation }}\left( 1 \right) \cr & \boxed{c = \frac{3}{2}} \cr & \therefore \,\boxed{cd = \frac{3}{2}} \cr} $$

$$\eqalign{ & {x^2} - 4x + 1 = 0 \cr & x + \frac{1}{x} = 4 \cr & {x^2} + \frac{1}{{{x^2}}} = 14 \cr & {x^4} + \frac{1}{{{x^4}}} = 194 \cr & {x^9} + {x^7} - 194{x^5} - 194{x^3} \cr & {x^9} + {x^7} - \left( {{x^4} + \frac{1}{{{x^4}}}} \right){x^5} - \left( {{x^4} + \frac{1}{{{x^4}}}} \right){x^3} \cr & = {x^9} + {x^7} - {x^9} - x - {x^7} - \frac{1}{x} \cr & = - \left( {x + \frac{1}{x}} \right) \cr & = - 4 \cr} $$

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

x + y = 3
Put x = 1, y = 2
x³ + y³ + 9xy
= (1)³ + (2)³ + 9 × 2 × 1
= 1 + 8 + 18
= 27

Alternate:
x + y = 3 . . . . . . (1)
On taking cube on both sides
(x + y)³ = (3)³
x³ + y³ + 3xy(x + y) = 27
x³ + y³ + 3xy(3) = 27
x³ + y³ + 9xy = 27

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 7 \cr & x + \frac{1}{x} = 3 \cr & {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 = 18 \cr} $$

$$\eqalign{ & x = 5 + 2\sqrt 6 \cr & \Rightarrow x = {\left( {\sqrt 3 + \sqrt 2 } \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 3 + \sqrt 2 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 3 - \sqrt 2 \cr & \Rightarrow \sqrt x + \frac{1}{{\sqrt x }} = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & \Rightarrow \sqrt x + \frac{1}{{\sqrt x }} = 2\sqrt 3 \cr} $$

$$\eqalign{ & {x^4} + {x^{ - 4}} = 194 \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = 194 + 2 \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = 196 \cr & {x^2} + \frac{1}{{{x^2}}} = 14 \cr & x + \frac{1}{x} = {\left( {16} \right)^{\frac{1}{2}}} \cr & x + \frac{1}{x} = 4 \cr & x + \frac{1}{x} + 2 = 4 + 2 \cr & x + \frac{1}{x} = 6 \cr} $$

(a + b)³ = a³ + 3a²b + 3ab² + b³ . . . . . . . . (i)
(a + 4)³ = a³ + 12a² + ka + 64
= (a)³ + 3.a².4 + k.a + (4)³ . . . . . . . . (ii)
Comparing equation (i) and (ii) we get,
b = 4 and k = 3k²
∴ k = 3 × 4² = 48

$$\eqalign{ & \left( {135\sqrt 5 {x^3} - 2\sqrt 2 {y^3}} \right) \div \left( {3\sqrt 5 x - \sqrt 2 y} \right) = A{x^2} + B{y^2} + \sqrt {10} Cxy \cr & \Rightarrow \frac{{{{\left( {3\sqrt 5 x} \right)}^3} - {{\left( {\sqrt 2 y} \right)}^3}}}{{3\sqrt 5 x - \sqrt 2 y}} = A{x^2} + B{y^2} + \sqrt {10} Cxy \cr & \Rightarrow \frac{{\left( {3\sqrt 5 x - \sqrt 2 y} \right)\left( {45{x^2} + 2{y^2} + 3\sqrt {10} xy} \right)}}{{\left( {3\sqrt 5 x - \sqrt 2 y} \right)}} = A{x^2} + B{y^2} + \sqrt {10} Cxy \cr & {\text{Comparison both side,}} \cr & A = 45,\,B = 2,\,C = 3 \cr & A + B - 9C = \left( {45 + 2 - 27} \right) \cr & A + B - 9C = 20 \cr} $$