Solutions to Challenging Algebra Questions
The solutions to the challenging algebra questions are presented along with detailed explanations.
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Solutions
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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
\( x^5 + \dfrac{1}{x^5} = 18 \sqrt 3 \times 10 - 2 \sqrt 3 = 178 \sqrt 3 \)
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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
\( (1-i)^{1+i} = \sqrt 2 \; e^{\frac{\pi}{4}} \cos (\ln \sqrt 2 - \frac{\pi}{4}) + i \; \sqrt 2 \; e^{\frac{\pi}{4}} \sin (\ln \sqrt 2 - \frac{\pi}{4}) \)
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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 \( k = 0, \pm 1, \pm 2, ... \)
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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: \( \{1/6 , 1/3 , 1\} \)
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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
\( x^4 + y^4 = \dfrac{584}{9} \)
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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
\( \left(\dfrac{2+\sqrt{-4}}{2}\right)^{10} + \left(\dfrac{2-\sqrt{-4}}{2}\right)^{10} = 0 \)
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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: \( \{ 2 , -\dfrac{3}{2} + i\dfrac{\sqrt{19}}{2} , -\dfrac{3}{2} - i\dfrac{\sqrt{19}}{2} \} \)
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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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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
\( \quad \dfrac{a\sqrt{b^2} - b \sqrt {a^2}}{\sqrt{(ab)^2}} \)
\( \quad \quad = \dfrac{- ab - ba }{-ab} = 2\)