\(x^{4} - 5x^{2} + 4 = 0\) looks like a quartic, but it is really a quadratic in \(x^{2}\). Spot the hidden quadratic, substitute, solve, then undo the substitution and throw away the values that are impossible.
If an equation has the shape \(a\big[f(x)\big]^{2} + b\,f(x) + c = 0\), put \(u = f(x)\), solve \(au^{2} + bu + c = 0\), then solve \(f(x) = u\) for each root. Reject any value that \(f(x)\) can never take, such as a negative \(x^{2}\) or a negative \(\sqrt{x}\).
Fig. 1: y = x⁴ - 5x² + 4 = (x² - 1)(x² - 4) meets the x-axis four times, at x = ±2 and x = ±1
Section 01
Spot the hidden quadratic.
Look for two terms where one power is exactly double the other, plus a constant. The smaller power is your new letter \(u\).
Equation contains
Let
Because
\(x^{4}\) and \(x^{2}\)
\(u = x^{2}\)
\(x^{4} = u^{2}\)
\(x\) and \(\sqrt{x}\)
\(u = \sqrt{x}\)
\(x = u^{2}\)
\(\dfrac{1}{x^{2}}\) and \(\dfrac{1}{x}\)
\(u = \dfrac{1}{x}\)
\(\dfrac{1}{x^{2}} = u^{2}\)
\(x^{\frac{2}{3}}\) and \(x^{\frac{1}{3}}\)
\(u = x^{\frac{1}{3}}\)
\(x^{\frac{2}{3}} = u^{2}\)
\(\sin^{2} x\) and \(\sin x\)
\(u = \sin x\)
\(\sin^{2} x = u^{2}\)
Worked example Four roots
Solve \(x^{4} - 5x^{2} + 4 = 0\).
Let \(u = x^{2}\): \(u^{2} - 5u + 4 = 0\).
\((u - 1)(u - 4) = 0\), so \(u = 1\) or \(u = 4\).
Let \(u = x^{2}\): \(u^{2} + 3u - 10 = 0 \Rightarrow (u + 5)(u - 2) = 0\).
\(u = -5\) is impossible, because \(x^{2}\) is never negative for real \(x\).
\(u = 2\): \(x^{2} = 2\), so \(x = \pm\sqrt{2}\).
Answer: \(x = \sqrt{2}\) or \(x = -\sqrt{2}\)
Examiner trap
Stopping at \(u = 1\) or \(u = 4\) answers the wrong question. The equation was in \(x\), so convert back, and remember that \(x^{2} = 4\) has TWO solutions, \(\pm 2\).
Section 02
Roots and reciprocals.
With \(\sqrt{x}\), remember that the symbol means the positive square root, so a negative value of \(u = \sqrt{x}\) must be rejected. With reciprocals, \(u = \dfrac{1}{x}\) can be any value except 0.
Worked example Square root
Solve \(6x - \sqrt{x} - 2 = 0\).
Let \(u = \sqrt{x}\), so \(x = u^{2}\): \(6u^{2} - u - 2 = 0\).
\((3u - 2)(2u + 1) = 0\), so \(u = \dfrac{2}{3}\) or \(u = -\dfrac{1}{2}\).
\(\sqrt{x} = -\dfrac{1}{2}\) is impossible, so reject it.
\(\sqrt{x} = \dfrac{2}{3} \Rightarrow x = \left(\dfrac{2}{3}\right)^{2}\).
\((2u - 1)(u - 3) = 0\), so \(u = \dfrac{1}{2}\) or \(u = 3\).
\(\dfrac{1}{x} = \dfrac{1}{2} \Rightarrow x = 2\); \(\dfrac{1}{x} = 3 \Rightarrow x = \dfrac{1}{3}\).
Answer: \(x = 2\) or \(x = \dfrac{1}{3}\)
Examiner trap
Check answers in the ORIGINAL equation. For \(6x - \sqrt{x} - 2 = 0\), \(x = \frac{1}{4}\) (from \(u = -\frac{1}{2}\)) gives \(1.5 - 0.5 - 2 = -1 \ne 0\), which confirms it must be rejected.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which substitution turns \(x^{4} - 7x^{2} + 12 = 0\) into a quadratic?
Quick check+10 XP
With \(u = x^{2}\), an equation becomes \(u^{2} - 13u + 36 = 0\). Solve it for \(x\).
Quick check+10 XP
Solving \(x + \sqrt{x} - 6 = 0\) gives \(\sqrt{x} = 2\) or \(\sqrt{x} = -3\). So \(x =\)
Quick check+10 XP
Using \(u = \dfrac{1}{x}\), the equation \(\dfrac{3}{x^{2}} + \dfrac{2}{x} - 1 = 0\) becomes
Section 03
Other functions of x.
The same idea works for any function: fractional powers, powers of a number and trigonometric functions. Cube roots can be negative, but \(\sin x\) and \(\cos x\) must lie between \(-1\) and \(1\).
Let \(u = x^{\frac{1}{3}}\): \(u^{2} - 3u + 2 = 0 \Rightarrow (u - 1)(u - 2) = 0\).
\(x^{\frac{1}{3}} = 1 \Rightarrow x = 1^{3} = 1\).
\(x^{\frac{1}{3}} = 2 \Rightarrow x = 2^{3} = 8\).
Answer: \(x = 1\) or \(x = 8\)
Worked example Trigonometric
Solve \(2\sin^{2} x - \sin x - 1 = 0\) for \(0^{\circ} \le x \le 360^{\circ}\).
Let \(u = \sin x\): \(2u^{2} - u - 1 = 0 \Rightarrow (2u + 1)(u - 1) = 0\).
\(\sin x = 1 \Rightarrow x = 90^{\circ}\).
\(\sin x = -\dfrac{1}{2}\): the reference angle is \(30^{\circ}\) and sine is negative in the third and fourth quadrants, so \(x = 180^{\circ} + 30^{\circ}\) or \(360^{\circ} - 30^{\circ}\).
Let \(u = 2^{x}\), so \(2^{2x} = u^{2}\): \(u^{2} - 6u + 8 = 0 \Rightarrow u = 2\) or \(4\). \(2^{x} = 2 \Rightarrow x = 1\); \(2^{x} = 4 \Rightarrow x = 2\).
Examiner trap
Do not reject a negative value of a cube root: \(x^{\frac{1}{3}} = -4\) is fine and gives \(x = -64\). Only reject values the function can never take (negative \(\sqrt{x}\) or \(x^{2}\), \(\sin x\) outside \(-1\) to \(1\), negative \(2^{x}\)).
Stage 2 · Exam
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