Math Practice
Twelve hand-built equations in quadratic form with full worked solutions. Each shows the choice of u, the quadratic in u, its solutions, and the back-substitution that recovers every value of the original variable.
Solve \(x^{4} - 5x^{2} + 4 = 0\).
\(u = x^{2}\): \((u - 1)(u - 4) = 0\), so \(u = 1\) or \(u = 4\). Then \(x^{2} = 1 \Rightarrow x = \pm 1\); \(x^{2} = 4 \Rightarrow x = \pm 2\).
Answer: \(x = \pm 1, \pm 2\).
Solve \(x^{4} - 13x^{2} + 36 = 0\).
\(u = x^{2}\): \((u - 4)(u - 9) = 0\). \(x = \pm 2, \pm 3\).
Solve \(x^{4} - 10x^{2} + 9 = 0\).
\(u = x^{2}\): \((u - 1)(u - 9) = 0\). \(x = \pm 1, \pm 3\).
Solve \(x^{4} - 3x^{2} - 4 = 0\).
\(u = x^{2}\): \((u - 4)(u + 1) = 0\). \(x^{2} = 4 \Rightarrow x = \pm 2\); \(x^{2} = -1 \Rightarrow x = \pm i\).
Solve \(x - 6\sqrt{x} + 8 = 0\).
\(u = \sqrt{x}\): \((u - 2)(u - 4) = 0\), so \(u = 2\) or \(u = 4\). Then \(x = 4\) or \(x = 16\). Both check.
Solve \(x^{2/3} - 4x^{1/3} + 3 = 0\).
\(u = x^{1/3}\): \((u - 1)(u - 3) = 0\), so \(u = 1\) or \(u = 3\). Then \(x = 1\) or \(x = 27\).
Solve \((x - 3)^{2} - 5(x - 3) + 6 = 0\).
\(u = x - 3\): \((u - 2)(u - 3) = 0\), so \(u = 2\) or \(u = 3\). Then \(x = 5\) or \(x = 6\).
Solve \(x^{6} - 7x^{3} - 8 = 0\).
\(u = x^{3}\): \((u - 8)(u + 1) = 0\), so \(u = 8\) or \(u = -1\). Then \(x^{3} = 8 \Rightarrow x = 2\); \(x^{3} = -1 \Rightarrow x = -1\).
Solve \(x - 5\sqrt{x} - 6 = 0\).
\(u = \sqrt{x}\): \((u - 6)(u + 1) = 0\), so \(u = 6\) or \(u = -1\). Reject \(u = -1\) (a square root can't be negative). \(u = 6 \Rightarrow x = 36\). Check: \(36 - 30 - 6 = 0\). ✓
Solve \((2x + 1)^{2} - 3(2x + 1) - 10 = 0\).
\(u = 2x + 1\): \((u - 5)(u + 2) = 0\), so \(u = 5\) or \(u = -2\). Then \(2x + 1 = 5 \Rightarrow x = 2\); \(2x + 1 = -2 \Rightarrow x = -\tfrac{3}{2}\).
Solve \(2x^{4} - 5x^{2} + 2 = 0\).
\(u = x^{2}\): \(2u^{2} - 5u + 2 = 0 \Rightarrow (2u - 1)(u - 2) = 0\), so \(u = 2\) or \(u = \tfrac{1}{2}\). \(x = \pm\sqrt{2}\) and \(x = \pm\tfrac{1}{\sqrt{2}}\).
Solve \((x^{2} - 2x)^{2} - 11(x^{2} - 2x) + 24 = 0\).
\(u = x^{2} - 2x\): \((u - 3)(u - 8) = 0\), so \(u = 3\) or \(u = 8\).
\(x^{2} - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0 \Rightarrow x = 3, -1\).
\(x^{2} - 2x - 8 = 0 \Rightarrow (x - 4)(x + 2) = 0 \Rightarrow x = 4, -2\).
Answer: \(x = -2, -1, 3, 4\).
The Equations Reducible to Quadratic Form lesson covers three substitution patterns; this set is 12 equations to practice on, each fully solved.
These problems are hand-written, not auto-generated. Identify the repeated expression, substitute
Stopping at the
Forgetting the
Discarding a negative
Skipping the extraneous check on equations with
The quadratic in
It can be written as
Let
Not always. Each
Whenever the substitution or back-substitution uses an even root — like
Back-substitute: replace
Several answers involve radicals or multiple values, which an automatic checker can't verify. Each problem has a full solution instead.