Solutions to Challenging Algebra Questions

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

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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 \( a = - x^2 - x/2 \)
    \( 2 a^2 + a (4x^2-x) + 2x^4 - x^3 - x^2 = K (a + x^2 + x) (a + x^2 + x/2) \)
    \( K (a + x^2 - x) (a + x^2 + x/2) = 0 \)
    \( a + x^2 - x = 0 \)
    \( x = \dfrac{1 \pm \sqrt{1-4a}}{2} \)
    \( a + x^2 + x/2 = 0 \)
    \( 2 a + 2x^2 + x = 0 \)
    \( x = \dfrac{ - 1 \pm \sqrt{1-16a}}{4} \)


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