Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

a = 1 + √3
∴ a² = 1 + 3 + 2√3
b = 1 - √3
∴ b² = 1 + 3 - 2√3
∴ a² + b² = 8

$$\eqalign{ & A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}} \cr & = \frac{{{{\left( {0.6} \right)}^3} + {{\left( {0.2} \right)}^3}}}{{{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)}} \cr & = \frac{{\left( {0.6 + 0.2} \right)\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}}{{\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}} \cr & = 0.8 \cr & B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}} \cr & = \frac{{{{\left( {0.9} \right)}^3} - {{\left( {0.3} \right)}^3}}}{{{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)}} \cr & = \frac{{\left( {0.9 - 0.3} \right)\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}}{{\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}} \cr & = 0.6 \cr & \therefore \,{\left( {{A^2} + {B^2}} \right)^2} \cr & = {\left[ {{{\left( {0.8} \right)}^2} + {{\left( {0.6} \right)}^2}} \right]^2} \cr & = {\left( {0.64 + 0.36} \right)^2} \cr & = {\left( 1 \right)^2} \cr & = 1 \cr} $$

$$\eqalign{ & 2{x^2} - 7x + 5 = 0 \cr & 2x + \frac{5}{x} = 7 \cr & x + \frac{5}{{2x}} = \frac{7}{2} \cr & {\left( {x + \frac{5}{{2x}}} \right)^2} = {\left( {\frac{7}{2}} \right)^2} \cr & {x^2} + \frac{{25}}{{4{x^2}}} + 2 \times x \times \frac{5}{{2x}} = \frac{{49}}{4} \cr & {x^2} + \frac{{25}}{{4{x^2}}} = \frac{{49}}{4} - 5 \cr & {x^2} + \frac{{25}}{{4{x^2}}} = \frac{{29}}{4} \cr & {x^2} + \frac{{25}}{{4{x^2}}} = 7\frac{1}{4} \cr} $$

4pxy = (x + 2y)² - (x - 2y)²
⇒ 4pxy = x² + 4y² + 4xy - x² - 4y² + 4xy
⇒ 4pxy = 8xy
⇒ $$p = \frac{{8xy}}{{4xy}} = 2$$

Alternate:
(a + b)² - (a - b)² = 4ab
4pxy = 4 × x × 2y
4pxy = 8xy
p = 2

$$\eqalign{ & 3{x^2} - 9x + 3 = 0 \cr & 3x\left( {x - 3 + \frac{1}{x}} \right) = 0,\,x + \frac{1}{x} = 3 \cr & \therefore \,{\left( {x + \frac{1}{x}} \right)^3} = {\left( 3 \right)^3} = 27 \cr} $$

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

$$\eqalign{ & 2{x^2} + 5x + 1 = 0 \cr & 2{x^2} + 1 = - 5x \cr & 2x + \frac{1}{x} = - 5 \cr & {\text{divide '2' both sides}} \cr & x + \frac{1}{{2x}} = - \frac{5}{2} \cr & a - b = \sqrt {{{\left( {a + b} \right)}^2} - 4ab} \cr & x - \frac{1}{{2x}} = \sqrt {{{\left( { - \frac{5}{2}} \right)}^2} - 4 \times x \times \frac{1}{{2x}}} \cr & = \sqrt {\frac{{25}}{2} - 2} \cr & = \frac{{\sqrt {17} }}{2} \cr} $$

$$\eqalign{ & {x^4} - 6{x^2} - 1 = 0, \cr & {x^2}\left( {{x^2} - 6 - \frac{1}{{{x^2}}}} \right) = 0 \cr & {x^2} - \frac{1}{{{x^2}}} = 6 \cr & {x^6} - \frac{1}{{{x^6}}} = 216 + 18 = 234 \cr & {x^6} - \frac{1}{{{x^6}}} - 5\left( {{x^2} - \frac{1}{{{x^2}}}} \right) + 5 \cr & = 234 - 30 + 5 \cr & = 209 \cr} $$

Given,
a + b + c = -11
By putting value i.e.
a = -4
b = -5
c = -2
Therefore,
(a + 4)³ + (b + 5)³ + (c + 2)³ - 3(a + 4)(b + 5)(c + 2)
= 0 + 0 + 0 - 0
= 0

$$\eqalign{ & {\text{Given, }}x - \frac{1}{x} = 5 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = {5^3} + 3 \times 5 = 140 \cr & \frac{{{x^6} + 3{x^3} - 1}}{{{x^6} - 8{x^3} - 1}} \cr & = \frac{{{x^3} + 3 - \frac{1}{{{x^3}}}}}{{{x^3} - 8 - \frac{1}{{{x^3}}}}} \cr & = \frac{{{x^3} - \frac{1}{{{x^3}}} + 3}}{{{x^3} - \frac{1}{{{x^3}}} - 8}} \cr & = \frac{{140 + 3}}{{140 - 8}} \cr & = \frac{{143}}{{132}} \cr & = \frac{{13}}{{12}} \cr} $$