Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \left( {0.4x + \frac{1}{x}} \right) = 5 \cr & {\text{cube both side}} \cr & {\left( {0.4x} \right)^3} + \frac{1}{{{x^3}}} + 3\left( {0.4x + \frac{1}{x}} \right) \times \left( {0.4x} \right)\left( {\frac{1}{x}} \right) = {\left( 5 \right)^3} \cr & 0.064{x^3} + \frac{1}{{{x^3}}} + 3\left( 5 \right)\left( {0.4} \right) = 125 \cr & 0.064{x^3} + \frac{1}{{{x^3}}} + 6 = 125 \cr & 0.064{x^3} + \frac{1}{{{x^3}}} = 119 \cr} $$

Given, x + y + z = 0
By putting values x = 2, y = -1, z = -1
$$\eqalign{ & {\text{Now, }}\frac{{xy + yz + zx}}{{{x^2} + {y^2} + {z^2}}} \cr & = \frac{{ - 2 + 1 - 2}}{{4 + 1 + 1}} \cr & = \frac{{ - 1}}{2} \cr} $$

$$\eqalign{ & P = 7 + 4\sqrt 3 \cr & PQ = 1,\,Q = \frac{1}{P} \cr & Q = \frac{1}{{7 + 4\sqrt 3 }} = 7 - 4\sqrt 3 \cr & P + Q \cr & = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 \cr & = 14 \cr & \Rightarrow \frac{1}{{{P^2}}} + \frac{1}{{{Q^2}}} \cr & = \frac{{{P^2} + {Q^2}}}{{{{\left( {PQ} \right)}^2}}} \cr & = \frac{{{{\left( {P + Q} \right)}^2} - 2PQ}}{{{{\left( {PQ} \right)}^2}}} \cr & = {\left( {14} \right)^2} - 2 \cr & = 196 - 2 \cr & = 194 \cr} $$

a = 55
b = 17
c = -72
a + b + c = 55 + 17 - 72 = 0
∴ a³ + b³ + c³ - 3abc = 0
(a + b + c) = 0
Answer = 0

(a - b) = 4
ab = 2
(a³ - b³)
= (a - b)[(a - b)² + 3ab]
= 4[4² + 3 × 2]
= 4[16 + 6]
= 88

Given:
x + y + z = 19, x² + y² + z² = 133 and xz = y²,
Formula used:
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
Calculation:
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
⇒ (19)² = 133 + 2(xy + yz + y²)
⇒ 133 + 2[y(x + y + z)] = 361
⇒ 2y(19) = 361 - 133
⇒ y = 6
x + y + z = 19
⇒ x + z = 13
The possible value of x and z is 9 and 4
x - 4
⇒ 9 - 4
⇒ 5
∴ The value is 5

$$a - \frac{1}{{a - 5}} = 10$$
We can write it is as (subtracting 5 from both sides)
$$\eqalign{ & a - 5 - \frac{1}{{a - 5}} = 10 - 5 \cr & {\text{Now, }}\left( {a - 5} \right) - \frac{1}{{\left( {a - 5} \right)}} = 5 \cr} $$
So, take the cube of this equation,
$$\eqalign{ & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 125 + 3 \times 5 \cr & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 140 \cr} $$

$$\eqalign{ & {x^2} - 16x + 59 = 0 \cr & {\left( {x - 6} \right)^2} + \frac{1}{{{{\left( {x - 6} \right)}^2}}} = ? \cr & {\text{Let }}x - 6 = t \cr & x = t + 6 \cr & {t^2} + \frac{1}{{{t^2}}} = ? \cr & {\left( {t + 6} \right)^2} - 16\left( {t + 6} \right) + 59 = 0 \cr & {t^2} + 12t - 36 - 16t - 96 + 59 = 0 \cr & {t^2} - 4t - 1 = 0 \cr & {\text{Dividing by }}t,{\text{ we get}} \cr & t - 4 - \frac{1}{t} = 0 \cr & t - \frac{1}{t} = 4 \cr & {\text{By squaring, we get}} \cr & {t^2} + \frac{1}{{{t^2}}} - 2 = 16 \cr & {t^2} + \frac{1}{{{t^2}}} = 18 \cr} $$

$$\eqalign{ & x = \sqrt 5 + 1,\,y = \sqrt 5 - 1 \cr & \frac{{{x^2}}}{{{y^2}}} + \frac{{{y^2}}}{{{x^2}}} + 4\left[ {\frac{x}{y} + \frac{y}{x}} \right] + 6 \cr & = {\left[ {\frac{x}{y} + \frac{y}{x}} \right]^2} - 2 + 4\left[ {\frac{x}{y} + \frac{y}{x}} \right] + 6 \cr & = {\left[ {\frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}} \right]^2} + 4\left[ {\frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}} \right] + 4 \cr & = {\left[ {\frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}}} \right]^2} + 4\left[ {\frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}}} \right] + 4 \cr & = {\left[ {\frac{{12}}{4}} \right]^2} + 4\left[ {\frac{{12}}{4}} \right] + 4 \cr & = {\left( 3 \right)^2} + 4\left( 3 \right) + 4 \cr & = 9 + 12 + 4 \cr & = 25 \cr} $$

$$\eqalign{ & {a^3} + 3{a^2} + 9a = 1.......\left( {\text{i}} \right) \cr & {a^3} + \frac{3}{a} = ? \cr & {\text{Multiply by '}}a{\text{' and '3' in equation}}\left( {\text{i}} \right) \cr & \left( {{a^3} + 3{a^2} + 9a = 1} \right) \times a \cr & \left( {{a^3} + 3{a^2} + 9a = 1} \right) \times 3 \cr & {a^4} + 3{a^3} + 9{a^2} = 1 \times a \cr & \underline {3{a^3} + 9{a^2} + 27a = 1 \times 3} \to \left( {{\text{Subtracting}}} \right) \cr & {a^4} - 27a = a - 3 \cr & {a^4} + 3 = 28a \cr & {\text{On dividing by }}a \cr & {a^3} + \frac{3}{a} = 28 \cr} $$