8.2 Equations Reducible to Quadratic Form
An equation that is not quadratic can often be made quadratic by a substitution. The signal to look for is an equation in which one power is the square of another — \(x^{4}\) and \(x^{2}\), or \(x\) and \(\sqrt {x}\), or \(e^{2x}\) and \(e^{x}\).
Note 8.5. The method is always the same three steps: give the repeated expression a name \(u\), solve the resulting quadratic in \(u\), then translate every value of \(u\) back into \(x\). The last step is where solutions are gained and lost, because one value of \(u\) may give two values of \(x\), one, or none at all.
Solution. Put \(u=x^{2}\), so that \(x^{4}=u^{2}\) and the equation becomes \[u^{2}-13u+36=0 \qquad \implies \qquad (u-4)(u-9)=0,\] so \(u=4\) or \(u=9\). Translating back, \[x^{2}=4 \implies x=\pm 2, \qquad x^{2}=9 \implies x=\pm 3 .\] The equation has four solutions, \(x=-3,-2,2,3\). Each value of \(u\) produced two values of \(x\) here because both were positive.
Solution. Put \(u=\sqrt {x}\), so that \(x=u^{2}\) and \[u^{2}-5u+6=0 \qquad \implies \qquad (u-2)(u-3)=0,\] giving \(u=2\) or \(u=3\). Then \(\sqrt {x}=2\) gives \(x=4\), and \(\sqrt {x}=3\) gives \(x=9\). Both check in the original equation.
Note 8.8. Here \(u=\sqrt {x}\) cannot be negative, so a negative root of the quadratic in \(u\) would simply have been discarded — no value of \(x\) corresponds to it. That is the reverse of the previous example, where each \(u\) gave two \(x\). Always ask what values \(u\) is allowed to take before translating back.
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