# Solutions to Challenging Algebra Questions

The solutions to the challenging algebra questions are presented along with detailed explanations.



## Solutions

1. Solution to Question 1
Expand the square
$\left(x+\dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2 x \dfrac{1}{x}$ Simplify the term $2 x \dfrac{1}{x}$ to $2$
$= x^2 + \dfrac{1}{x^2} + 2$
Use $x^2 + \dfrac{1}{x^2} = 10$ to simplify the above and obtain
$\left(x+\dfrac{1}{x}\right)^2 = 10 + 2 = 12$
Take the square root of both sides
$\left(x+\dfrac{1}{x}\right) = \sqrt {12} = 2 \sqrt 3$
One way to obtain terms with $x^3$ and $\dfrac{1}{x^3}$, expand the following
$\left(x+\dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2}\right) = x^3 + \dfrac{1}{x} + x + \dfrac{1}{x^3}$
Deduce $x^3 + \dfrac{1}{x^3}$ from the above in terms of known quantities
$x^3 + \dfrac{1}{x^3} = \left(x+\dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2} \right) - \left(x+\dfrac{1}{x} \right)$
Substitute the known quantities by their numerical values
$x^3 + \dfrac{1}{x^3} = 2 \sqrt {3} \times 10 - 2 \sqrt {3} = 18 \sqrt 3$
One way to obtain terms with $x^5$ and $\dfrac{1}{x^5}$, expand the following
$\left( x^3 + \dfrac{1}{x^3} \right) \left( x^2 + \dfrac{1}{x^2} \right) = x^5 + x + \dfrac{1}{x} + \dfrac{1}{x^5}$
Deduce $x^5 + \dfrac{1}{x^5}$ from the above in terms of known quantities
$x^5 + \dfrac{1}{x^5} = \left( x^3 + \dfrac{1}{x^3} \right) \left( x^2 + \dfrac{1}{x^2} \right) - \left(x + \dfrac{1}{x} \right)$
Substitute known quantities by their numerical values
$x^5 + \dfrac{1}{x^5} = 18 \sqrt 3 \times 10 - 2 \sqrt 3 = 178 \sqrt 3$

2. Solution to Question 2
Write the complex numbers $1-i$ in exponential form
$1 - i = \sqrt 2 \left( \frac{\sqrt 2}{2} + i \frac{- \sqrt 2}{2} \right) = \sqrt 2 e^{- i \frac{\pi}{4}}$
Substitute $1-i$ by the above in the given expression $(1-i)^{1+i}$
$(1-i)^{1+i} = \left(\sqrt 2 e^{- i \frac{\pi}{4}} \right)^{1+i}$
Use exponents rule to take $\sqrt 2$ from inside the brackets
$(1-i)^{1+i} = (\sqrt 2) ^{1+i} \left(e^{ - i \frac{\pi}{4} } \right)^{(1+i)}$
Use exponents rule to rewrite as
$(1-i)^{1+i} = \sqrt 2 \; e^{\frac{\pi}{4}} \; \sqrt 2^{\; i} e^{- i \frac{\pi}{4}}$
$\sqrt 2$ may be written as $e^{\ln \sqrt 2}$ ; hence the given expression may be written as
$(1-i)^{1+i} = \sqrt 2 e^{\frac{\pi}{4}} (e^{\ln \sqrt 2})^i e^{- i \frac{\pi}{4}}$
Simplify
$(1-i)^{1+i} = \sqrt 2 \; e^{\frac{\pi}{4}} \; e^{ (\ln \sqrt 2 - \frac{\pi}{4})i}$
Write in standard form
$(1-i)^{1+i} = \sqrt 2 \; e^{\frac{\pi}{4}} \cos (\ln \sqrt 2 - \frac{\pi}{4}) + i \; \sqrt 2 \; e^{\frac{\pi}{4}} \sin (\ln \sqrt 2 - \frac{\pi}{4})$

3. Solution to Question 3
Let us rewrite the left side of the given expression in the form $R \cos (x + \pi/4 - \phi)$.
We first rewrite the right side as
$6 \cos(x + \pi/4) + 8 \sin(x + \pi/4)$
$= \sqrt{6^2+8^2} \left ( \dfrac{6}{\sqrt{6^2+8^2}} \cos(x + \pi/4) + \dfrac{8}{\sqrt{6^2+8^2}} \sin(x + \pi/4) \right)$
Since $\sqrt{6^2+8^2} = 10$ the above simplifies to
$= 10 \left ( 0.6 \cos(x + \pi/4) + 0.8 \sin(x + \pi/4) \right)$
Let $\cos \phi = \dfrac{6}{\sqrt{6^2+8^2}} = 0.6$ and $\sin \phi = \dfrac{8}{\sqrt{6^2+8^2}} = 0.8$
Check that $(\cos \phi)^2 + (\sin \phi)^2 = 0.6^2 + 0.8^2 = 1$
which gives $\phi = \arctan \left(\dfrac{8}{6} \right)$
We now write the right side of the given equation using angle $\phi$ as follows
$6 \cos(x + \pi/4) + 8 \sin(x + \pi/4)$
$= 10 \left( \cos \phi \cos(x + \pi/4) + \sin \phi \sin(x + \pi/4) \right)$
We use the trigonometric formula $\cos(A-B) = \cos A \cos B+\sin A \sin B$ to rewrite the above as
$6 \cos(x + \pi/4) + 8 \sin(x + \pi/4) = 10 \cos (x + \pi/4 - \phi)$
The given trigonometric equation may be written as
$10 \cos (x + \pi/4 - \phi) = 5$
$\cos (x + \pi/4 - \phi) = \dfrac{1}{2}$
which gives two sets of solutions
$x + \pi/4 - \phi = \pi/3 + 2 k \pi$
First solution set: $x = \phi + \pi/12 + 2 k \pi = \arctan(8/6) + \pi/12 + 2 k \pi$ for $k = 0, \pm 1, \pm 2, ...$
$x + \pi/4 - \phi = 5\pi/3 + 2 k \pi$
Second solution set: $x = \phi + 17 \pi/12 + 2 (k + 1) \pi = \arctan(8/6) + 17 \pi/12 + 2 (k + 1) \pi$ for $k = 0, \pm 1, \pm 2, ...$

4. Solution to Question 4
Use rational exponent in place of the square root on the left side of the equation
$(x^{\frac{1}{2}})^{|x|} = x^{ x^2+\frac{1}{18}}$
Use rule of exponents to rewrite that above as
$x^{\frac{1}{2}|x|} = x^{ x^2+\frac{1}{18}}$
Take the $\ln$ of both sides
$\ln ( x^{\frac{1}{2}|x|}) = \ln (x^{ x^2+\frac{1}{18}})$
Use the $\ln$ rule: $\ln x^y = y \ln x$ to rewrite the above as
$\frac{1}{2}|x| \ln (x) = (x^2+\frac{1}{18}) \ln (x)$
Put all terms on one side and factor
$\frac{1}{2}|x| \ln (x) - (x^2+\frac{1}{18}) \ln (x) = 0$
$\ln(x) (\frac{1}{2}|x| - x^2 - \frac{1}{18}) = 0$
The solution are found by setting each of the two factors above equal to 0.
1) Solve $\ln(x) = 0$ which gives the solution $x = 1$
2) Solve $\frac{1}{2}|x| - x^2 - \frac{1}{18} = 0$
Let $y = |x|$ and $y^2 = |x|^2 = x^2$
Substitute $|x|$ and $x^2$ in the equation 2) and rewrite it as a quadratic equation
$\frac{1}{2} y - y ^2 - \frac{1}{18} = 0$
Solve the above quadratic equation to obtain the solutions
$y = 1/6$ and $y = 1/3$
|x| = y = 1/6 gives x = 1/6 , $x$ must be positive because we are looking for real solutions only
|x| = y = 1/3 gives x = 1/3 , $x$ must be positive because we are looking for real solutions only
The solution set of the given eqaution is: $\{1/6 , 1/3 , 1\}$

5. Solution to Question 5
$x + y$ is a factor of $x^3 + y^3$ , hence division gives
$\dfrac{x^3 + y^3}{x+y} = x^2 - xy + y^2$     Eq-1
We also have
$\dfrac{x^3 + y^3}{x+y} = 24/4 = 6$     Eq-2
Combine Eq-1 and Eq-2 to write
$x^2 - xy + y^2 = 6$     Eq-3
Given $x + y = 4$, square both sides
$(x + y)^2 = 4^2$
Expand
$x^2 + y^2 + 2xy = 16$     Eq-4
Subtract Eq-3 and Eq-4 to write
$3xy = 10$
$x y = 10/3$
Use Eq-3 to write
$x^2 + y^2 = 6 + x y$
Substitute $x y$ by its value
$x^2 + y^2 = 6 + 10/3 = 28/3$
Square both sides
$(x^2 + y^2)^2 = 784/9$
Exapnd
$x^4 + y^4 + 2 x^2 y^2 = 784/9$
The above gives
$x^4 + y^4 = 784/9 - 2 x^2 y^2$
Rewrite as
$= 784/9 - 2 (xy)^2$
Substitute $x y$ by its numeraical value
$= 784/9 - 2 \left(\dfrac{10}{3} \right)^2$
Simplify
$x^4 + y^4 = \dfrac{584}{9}$

6. Solution to Question 6
Write the complex numbers $\dfrac{2+\sqrt{-4}}{2}$ and $\dfrac{2 - \sqrt{-4}}{2}$ in standard forms
$\dfrac{2+\sqrt{-4}}{2} = \dfrac{2 + 2\sqrt{- 1}}{2} = 1 + i$
$\dfrac{2 - \sqrt{-4}}{2} = \dfrac{2 - 2\sqrt{- 1}}{2} = 1 - i$
Hence
$\left(\dfrac{2+\sqrt{-4}}{2}\right)^{10} + \left(\dfrac{2-\sqrt{-4}}{2}\right)^{10} = (1+i)^{10} + (1-i)^{10}$
Write the complex numbers $1 + i$ and $1 - i$ in complex forms
$1 + i = \sqrt 2 e^{i \pi/4}$
$1 - i = \sqrt 2 e^{ - i \pi/4}$
Substitute $1 + i$ and $1 - i$ to obtain
$\left(\dfrac{2+\sqrt{-4}}{2}\right)^{10} + \left(\dfrac{2-\sqrt{-4}}{2}\right)^{10}$
$\quad \quad = (\sqrt 2 e^{i \pi/4})^{10} + (\sqrt 2 e^{ - i \pi/4})^{10}$
Use exponent rule $(e^x)^y = e^{xy}$
$= (\sqrt 2)^{10} ( e^{i 10 \pi/4} + e^{- i 10 \pi/4} )$
Use the fact that $10 \pi/4 = 2 \pi + \pi/2$ to rewrite the above as
$= (\sqrt 2)^{10} ( e^{ 2\pi i + i \pi/2} + e^{ - 2\pi i - i \pi/2})$
Use the exponent rule $e^{x+y} = e^x e^y$ to rewrite the above as
$= (\sqrt 2)^{10} ( e^{ 2\pi i} e^{i \pi/2} + e^{ - 2\pi i} e^{ - i \pi/2})$
Simplify using $e^{ 2\pi i} = 1$ and $e^{ - 2\pi i} = 1$
$= (\sqrt 2)^{10} ( e^{i \pi/2} + e^{ - i \pi/2})$
Note that
$e^{i \pi/2} + e^{ - i \pi/2} = i - i = 0$
Hence
$\left(\dfrac{2+\sqrt{-4}}{2}\right)^{10} + \left(\dfrac{2-\sqrt{-4}}{2}\right)^{10} = 0$

7. Solution to Question 7
Rewrite the given equation as follows
$(x - 2) + (x^2 - 4) + (x^3 - 8) = 0$
Note that each term between brackets has the factor $x - 2$
$x^2 - 4 = (x - 2)(x + 2)$
and using division, we have
$\dfrac{x^3 - 8}{x-2} = x^2 + 2x + 4$
which can also be writtean as
$(x^3 - 8) = (x - 2)(x^2 + 2x + 4)$
Hence the given equation may be written as
$(x - 2) + (x - 2)(x+2) + (x-2)(x^2 + 2x + 4) = 0$
Factor $x - 2$ out
$(x - 2) (1 + (x+2) + (x^2 + 2x + 4)) = 0$
Group like terms
$(x - 2) ( x^2 + 3x + 7) = 0$
$x - 2 = 0$ gives the solution $x = 2$
$x^2 + 3x + 7 = 0$ gives the complex solutions: $-\dfrac{3}{2} + i\dfrac{\sqrt{19}}{2}$ and $-\dfrac{3}{2} - i\dfrac{\sqrt{19}}{2}$
Solution set is: $\{ 2 , -\dfrac{3}{2} + i\dfrac{\sqrt{19}}{2} , -\dfrac{3}{2} - i\dfrac{\sqrt{19}}{2} \}$

8. Solution to Question 8
Let us rewrite the given equation as a quadratic equation in $a$
$2 a^2 + a (4x^2-x) + 2x^4 - x^3 - x^2 = 0$
Solve the above quadratic equation for $a$
Discriminant
$\Delta = (4x^2-x)^2 - 4(2)(2x^4 - x^3 - x^2) = 9x^2$
$a = \dfrac{-(4x^2-x) \pm \sqrt {9x^2 })}{4}$
Simplify the above solutions
$a = - x^2 + x$ and $a = - x^2 - x/2$
$2 a^2 + a (4x^2-x) + 2x^4 - x^3 - x^2 = K (a + x^2 + x) (a + x^2 + x/2)$
$K (a + x^2 - x) (a + x^2 + x/2) = 0$
$a + x^2 - x = 0$
$x = \dfrac{1 \pm \sqrt{1-4a}}{2}$
$a + x^2 + x/2 = 0$
$2 a + 2x^2 + x = 0$
$x = \dfrac{ - 1 \pm \sqrt{1-16a}}{4}$

9. Solution to Question 9
If $a \gt 0$ , then $|a| = a$ and If $b \lt 0$ , then $|b| = - b$
$\sqrt {x^2} |x|$
Therefore
$\sqrt{b^2} = |b| = - b$ , $\sqrt {a^2} = a$ and
$\sqrt{(ab)^2} = |ab| = |a| |b| = - a b$
Substitute all the above in the given expression and simplify
$\quad \dfrac{a\sqrt{b^2} - b \sqrt {a^2}}{\sqrt{(ab)^2}}$

$\quad \quad = \dfrac{- ab - ba }{-ab} = 2$
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