Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {t^2} - 4t + 1 = 0 \cr & \Rightarrow {t^2} + 1 = 4t \cr & \Rightarrow \frac{{{t^2} + 1}}{t} = \frac{{4t}}{t} \cr & \Rightarrow t + \frac{1}{t} = 4 \cr & \left[ {{\text{Take cube both sides}}} \right] \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3.t.\frac{1}{t}\left( {t + \frac{1}{t}} \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3\left( 4 \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 64 - 12 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 52 \cr} $$

$$\eqalign{ & \root 3 \of a + \root 3 \of b = \root 3 \of c \cr & {\text{Take cube both sides}} \cr & {\left( {\root 3 \of a + \root 3 \of b } \right)^3} = {\left( {\root 3 \of c } \right)^3} \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {\root 3 \of a + \root 3 \of b } \right) = c \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} = c \cr & \Rightarrow a + b - c = - 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again take cube both sides}} \cr & \Rightarrow {\left( {a + b - c} \right)^3} = - 27abc \cr & \Rightarrow {\left( {a + b - c} \right)^3} + 27abc = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }} \cr & a = 0 \cr & b = 1 \cr & c = 1 \cr & {\text{Value of }}{\left( {a + b - c} \right)^3} + 27abc \cr & = {\left( {0 + 1 - 1} \right)^3} + 27 \times 0 \times 1 \times 1 \cr & = 0 \cr} $$

 $${\text{Given}}\,x = \root 3 \of {a + \sqrt {{a^2} + {b^3}} } \,\,+ $$      $$\root 3 \of {a - \sqrt {{a^2} + {b^3}} } $$
$$\eqalign{ & {\text{Let}}\,z = \sqrt {{a^2} + {b^3}} \cr & \therefore x = \root 3 \of {a + z} + \root 3 \of {a - z} \cr & {\text{Cubing both side}} \cr} $$
  $$\therefore {x^3} = {\left( {\root 3 \of {a + z} } \right)^3} + {\left( {\root 3 \of {a - z} } \right)^3} \, + $$      $$\,3{\left( {a + z} \right)^{\frac{1}{3}}}$$  $${\left( {a - z} \right)^{\frac{1}{3}}} \times\, $$   $$\left( {\root 3 \of {a + z} + \root 3 \of {a - z} } \right)$$
 $$ \,\Rightarrow {x^3} = a + z + a - z \,+\, $$    $$3{\left( {{a^2} + az - az - {z^2}} \right)^{\frac{1}{3}}}$$    $$ \times \left( x \right)$$
$$\eqalign{ & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {z^2}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & {\text{Put the value of }}z \cr & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {a^2} - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a + 3{\left( { - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a - 3bx \cr & \therefore {x^3} + 3bx = 2a \cr} $$

$$\eqalign{ & \frac{{{x^{24}} + 1}}{{{x^{12}}}} = 7{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{{x^{24}}}}{{{x^{12}}}} + \frac{1}{{{x^{12}}}} = 7 \cr & \Rightarrow {x^{12}} + \frac{1}{{{x^{12}}}} = 7 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {\left( {{x^{12}} + \frac{1}{{{x^{12}}}}} \right)^3} = {7^3} \cr} $$
  $$ \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} + \frac{{3 \times {x^{12}} \times 1}}{{{x^{12}}}}$$     $$\left( {{x^{12}} + \frac{1}{{{x^{12}}}}} \right) = $$    $$343$$
$$\eqalign{ & \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} + 3\left( 7 \right) = 343 \cr & \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} = 343 - 21 \cr & \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} = 322 \cr & \Rightarrow \frac{{{x^{72}} + 1}}{{{x^{36}}}} = 322 \cr} $$

$$\eqalign{ & {\text{Given, }}x = 2 \cr & {\text{Find }}{x^3} + 27{x^2} + 243x + 631 \cr & {\text{To put value }}x = 2{\text{ }} \cr & \Rightarrow {2^3} + 27 \times {2^2} + \left( {243 \times 2} \right) + 631 \cr & \Rightarrow 8 + 108 + 486 + 631 \cr & \Rightarrow 1233 \cr} $$

5x + 9y = 5 . . . . . (i)
125x³ + 729y³ = 120 . . . . . (ii)
From equation (i) cubing both sides
⇒ (5x + 9y)³ = 5³
⇒ 125x³ + 729y³ + 3 × 5x × 9y(5x + 9y) = 125
⇒ 125x³ + 729y³ + 135xy × 5 = 125
⇒ 120 + 135xy × 5 = 125
⇒ 135xy × 5 = 5
⇒ xy = $$\frac{1}{{135}}$$
∴ product of x & y = $$\frac{1}{{135}}$$

$$\eqalign{ & {\text{Given,}} \cr & {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & {\text{Find, }}{x^3} + {y^3} + {z^3} = ? \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 2 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 1 - 1 \cr & \Rightarrow \left( {{x^2} + 1 - 2x} \right) + {y^2} + \left( {{z^2} + 1 - 2z} \right) = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} + {y^2} + {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} = 0 \cr & \Rightarrow x = 1 \cr & \Rightarrow {y^2} = 0 \cr & \Rightarrow y = 0 \cr & \Rightarrow {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow z = 1 \cr & {\text{Value substituted in question,}} \cr & \Rightarrow {x^3} + {y^3} + {z^3} \cr & \Rightarrow {1^3} + 0 + {1^3} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 1 \cr & {\text{Find }}\frac{2}{{{x^2} - x + 2}} = ? \cr & x + \frac{1}{x} = 1 \cr & {x^2} + 1 = x \cr & \left( {{x^2} - x} \right) = - 1 \cr & {\text{Putting value in,}} \cr & = \frac{2}{{{x^2} - x + 2}} \cr & = \frac{2}{{ - 1 + 2}} \cr & = 2 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ , }}y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & {\text{Find, }}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? \cr & \Rightarrow {\text{ }}\frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} \cr & \Rightarrow {\text{ }}\frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} = ? \cr & {\text{Now,}} \cr & {\text{ }}x + y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow x + y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} \cr & \Rightarrow {\text{ }}x + y = \frac{{2\left( {{{\sqrt 5 }^2} + {{\sqrt 3 }^2}} \right)}}{{5 - 3}} \cr & \Rightarrow x + y = 8\,.......(i) \cr & Again, \cr & x - y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}x - y = \frac{{4 \times \sqrt 5 \times \sqrt 3 }}{2} \cr & \Rightarrow x - y = 2\sqrt {15} ..............(ii) \cr & {\text{And, }}xy = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}xy = 1 \cr & {\text{Substitutes values in the question}}{\text{.}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} \cr & \Rightarrow \frac{{{8^2} - 1}}{{{{\left( {2\sqrt {15} } \right)}^2} + 1}} \cr & \Rightarrow \frac{{63}}{{61}} \cr} $$

$$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$              $$ ÷ $$  $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$    $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{ & \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr & \Rightarrow a = 1 + \frac{{10}}{{101}} \cr & \Rightarrow a = \frac{{111}}{{101}} \cr & \Rightarrow b = 1 - \frac{{10}}{{101}} \cr & \Rightarrow b = \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$