Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

3x + 4y - 2z + 9 = 17 . . . . . . . . (1)
7x + 2y + 11z + 8 = 23 . . . . . . . (2)
5x + 9y + 6z - 4 = 18 . . . . . . . . (3)
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15x + 15y + 15z + 13 = 58
(By adding equation (1), (2) & (3))
15(x + y + z) = 45
x + y + z = 3
x + y + z - 34 = 3 - 34 = -31

$$\eqalign{ & {x^4}\left( {{x^4} - 34 + \frac{1}{{{x^4}}}} \right) = 0 \cr & {x^4} + \frac{1}{{{x^4}}} = 34 \cr & {x^4} + \frac{1}{{{x^4}}} + 2 = 36 \cr & {x^2} + \frac{1}{{{x^2}}} = 6 \cr & x + \frac{1}{x} = \sqrt 8 \cr & {x^3} + \frac{1}{{{x^3}}} = 8\sqrt 8 - 3\sqrt 8 \cr & {x^3} + \frac{1}{{{x^3}}} = 5\sqrt 8 \cr} $$

$$\eqalign{ & {x^4} + \frac{1}{{{x^4}}} = \frac{{257}}{{16}} \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = \frac{{257}}{{16}} + 2 \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = {\left( {\frac{{17}}{4}} \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = \frac{{17}}{4} \cr & x + \frac{1}{x} = {\left( {\frac{{17}}{4} + 2} \right)^{\frac{1}{2}}} = \frac{5}{2} \cr & {\text{Here }}x = 2 \cr & {\text{Hence }}\frac{8}{{13}}\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = \frac{8}{{13}}\left( {8 + \frac{1}{8}} \right) \cr & = \frac{8}{{13}} \times \frac{{65}}{8} \cr & = 5 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3,\,x \ne 0 \cr & {x^2} + \frac{1}{{{x^2}}} = {3^2} - 2 = 7 \cr & {x^4} + \frac{1}{{{x^4}}} = {7^2} - 2 = 47 \cr & {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 \cr & = 27 - 9 \cr & = 18 \cr & {x^7} + \frac{1}{{{x^7}}} = \left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) - \left( {x + \frac{1}{x}} \right) \cr & = 47 \times 18 - 3 \cr & = 846 - 3 \cr & = 843 \cr} $$

a² + b² + c² = 6.25
ab + bc + ca = 0.52
As we know-
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b + c)² = 6.25 + 2 × 0.52
(a + b + c)² = 6.25 + 1.04
(a + b + c)² = 7.29
a + b + c = ±2.7
Since a + b + c < 0
then, a + b + c = -2.7

Concept used:
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - b³ - 3ab(a - b)
Calculation:
(x + y)³ - (x - y)³ - 3y(2x² - 3y²) = ky³
⇒ x³ + y³ + 3x²y + 3xy² - x³ + y³ + 3x²y - 3xy² - 6x²y + 9y³ = ky³
⇒ 11y³ = ky³
⇒ k = 11
∴ The value of K is 11

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

$$\eqalign{ & 5\sqrt 5 {x^3} + 2\sqrt 2 {y^3} = \left( {Ax + \sqrt 2 y} \right)\left( {B{x^2} + 2{y^2} + Cxy} \right) \cr & {\left( {\sqrt 5 x} \right)^3} + {\left( {\sqrt 2 y} \right)^3} = \left( {\sqrt 5 x + \sqrt 3 y} \right)\left( {5{x^2} + 9{y^2} - \sqrt {10} xy} \right) \cr & \left( {\sqrt 5 x + \sqrt 3 y} \right)\left( {5{x^2} + 9{y^2} - \sqrt {10} xy} \right) = \left( {Ax + \sqrt 2 y} \right)\left( {B{x^2} + 2{y^2} + Cxy} \right) \cr & {\text{Comparison both side:}} \cr & A = \sqrt 5 \cr & B = 5 \cr & C = - \sqrt {10} \cr & {A^2} + {B^2} + {C^2} = 5 + 25 - 10 = 20 \cr} $$

$$\eqalign{ & A = 1 + {2^P} \cr & B = 1 + {2^{ - P}} \cr & {\text{Put }}P = 1 \cr & A = 3,\,B = \frac{3}{2} \cr & {\text{Option C is correct}} \cr & \cr & {\bf{Alternate:}} \cr & A = 1 + {2^P}\,......\,\left( {\text{i}} \right) \cr & B = 1 + {2^{ - P}} \cr & = 1 + \frac{1}{{{2^P}}} \cr & = \frac{{{2^P} + 1}}{{{2^P}}} \cr & = \frac{A}{{A - 1}}\,\,\,\,\,\left( {{\text{from equation }}\left( {\text{i}} \right)} \right) \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 8 \cr & {x^2} + 1 = 8x \cr & \frac{{5x}}{{{x^2} + 1 - 6x}} \cr & = \frac{{5x}}{{8x - 6x}} \cr & = \frac{{5x}}{{2x}} \cr & = \frac{5}{2} \cr & = 2.5 \cr} $$