Challenging Algebra Questions with Solutions
Challenging algebra questions are presented along with their detailed solutions. These questions require the application of basic algebra rules, exponents, complex numbers, factoring, and logical thinking. Do not give up quickly; you will learn a lot while searching for the solution.
Question 1 — Algebraic Identities and Powers
If \( x^2 + \dfrac{1}{x^2} = 10 \), then what is the exact value of \( x^5 + \dfrac{1}{x^5} \) for \( x \ge 0 \)?
A) 9 B) 18 C) 1/18 D) 27 E) 178 \(\sqrt{3}\)
Show Step-by-Step Solution
\[ \left(x+\dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2 = 12 \implies x+\dfrac{1}{x} = 2\sqrt{3} \]
\[ x^3 + \dfrac{1}{x^3} = \left(x+\dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2} \right) - \left(x+\dfrac{1}{x} \right) = 18 \sqrt 3 \]
\[ x^5 + \dfrac{1}{x^5} = \left( x^3 + \dfrac{1}{x^3} \right) \left( x^2 + \dfrac{1}{x^2} \right) - \left(x + \dfrac{1}{x} \right) = \mathbf{178 \sqrt 3} \]
Question 2 — Complex Numbers & Exponents
Write the complex number \( (1-i)^{1+i} \) in standard form \( a + i b \).
A) Real Form B) Complex Exponential Form C) Logarithmic D) Polar E) None
Show Step-by-Step Solution
Convert to exponential form: \( 1-i = \sqrt 2 e^{- i \pi/4} \).
Substitute and expand: \( (\sqrt 2 e^{- i \pi/4} )^{1+i} = \sqrt 2 e^{\pi/4} e^{i (\ln \sqrt 2 - \pi/4)} \).
Result: \( \mathbf{\sqrt 2 e^{\frac{\pi}{4}} \cos \left(\ln \sqrt 2 - \frac{\pi}{4}\right) + i \sqrt 2 e^{\frac{\pi}{4}} \sin \left(\ln \sqrt 2 - \frac{\pi}{4}\right)} \).
Question 3 — Trigonometric Equations
Solve: \( 6 \cos(x + \pi/4) + 8 \sin(x + \pi/4) = 5 \).
A) Trig Identity Method B) Quadratic Formula C) Factoring D) Substitution E) None
Show Step-by-Step Solution
Transform to \( 10 \cos (x + \pi/4 - \phi) = 5 \), where \( \phi = \arctan(4/3) \).
\[ \cos (x + \pi/4 - \phi) = 1/2 \]
Two sets of solutions: \( x = \arctan(4/3) + \pi/12 + 2k\pi \) or \( x = \arctan(4/3) + 17\pi/12 + 2k\pi \).
Question 4 — Rational & Variable Exponents
Real solutions to: \( (\sqrt {x})^{|x|} = x^{ x^2+\frac{1}{18}} \)?
A) 0 B) -1 C) 0.5 D) 2 E) \{1/6, 1/3, 1\}
Show Step-by-Step Solution
Logarithmic reduction leads to: \( \ln(x) (\frac{1}{2}|x| - x^2 - \frac{1}{18}) = 0 \).
Factors give \( x=1 \) and quadratic solutions for \( |x| \): \( 1/6 \) and \( 1/3 \).
Final set: \( \mathbf{\{ 1/6, 1/3, 1 \}} \).
Question 5 — Polynomial Expansions
Given \( x + y = 4 \) and \( x^3 + y^3 = 24 \), evaluate \( x^4 + y^4 \).
A) 584/9 B) 60 C) 100 D) 24 E) 12
Show Step-by-Step Solution
Use \( x^3+y^3 = (x+y)(x^2-xy+y^2) \implies 6 = x^2-xy+y^2 \).
Calculate \( xy = 10/3 \) and \( x^2+y^2 = 28/3 \).
Calculate \( x^4+y^4 = (x^2+y^2)^2 - 2(xy)^2 = \mathbf{584/9} \).
Question 6 — Complex Number Simplification
Show: \( \left(\dfrac{2+\sqrt{-4}}{2}\right)^{10} + \left(\dfrac{2-\sqrt{-4}}{2}\right)^{10} = 0 \)
Show Step-by-Step Solution
Simplify to \((1+i)^{10} + (1-i)^{10}\).
Using polar form, this results in \( (\sqrt{2})^{10} (e^{i \pi/2} + e^{-i \pi/2}) \).
Since \( i + (-i) = 0 \), the total is \( \mathbf{0} \).
Question 7 — Polynomial Factoring
Solve: \( x + x^2 + x^3 = 2 + 4 + 8 \)
Show Step-by-Step Solution
Group: \( (x-2) + (x^2-4) + (x^3-8) = 0 \).
Factor out \((x-2)\): \( (x-2)(1 + x + 2 + x^2 + 2x + 4) = (x-2)(x^2 + 3x + 7) = 0 \).
Roots: \( x=2 \) and complex roots from the quadratic.
Question 8 — Multi-variable Quadratic
Solve for \( x \): \( 2x^4 - x^3 - x^2(1 - 4a) - ax + 2a^2 = 0 \)
Show Step-by-Step Solution
Treat as quadratic in \( a \): \( 2a^2 + a(4x^2 - x) + (2x^4 - x^3 - x^2) = 0 \).
Factor using the quadratic formula, yielding cases \( a = -x^2+x \) and \( a = -x^2-x/2 \).
Roots derived from: \( x^2 - x + a = 0 \) and \( 2x^2 + x + 2a = 0 \).
Question 9 — Absolute Values
Simplify \( \dfrac{a\sqrt{b^2} - b \sqrt {a^2}}{\sqrt{(ab)^2}} \) given \( a > 0, b < 0 \).
Show Step-by-Step Solution
Apply definitions: \(\sqrt{a^2}=a\), \(\sqrt{b^2}=-b\), \(\sqrt{(ab)^2}=-ab\).
Expression becomes \(\frac{a(-b) - b(a)}{-ab} = \frac{-2ab}{-ab} = \mathbf{2}\).