Problem

Solve \(x^4~-~5x^2~+~4~=~0\) by rewriting it as a quadratic in \(x^2\). Give the complete real solution set and check the values in the original equation.

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What this problem is really about

The repeated powers make this look like a quadratic in disguise. Substitute a single variable for the squared expression, factor the resulting quadratic, and then translate each result back to the original variable. Remember that undoing a square can produce two real values, and check them in the original equation.

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Let u=x². Then u²−5u+4=(u+1)(u+4)=0, so x²=−1 or −4 and there are no real solutions. The original equation is always positive.

Let u=x². Then u²−5u+4=(u−1)(u−4)=0, so x²=1 or 4 and x=1,2. These are only the positive original-equation solutions.

Let u=x². Then u²−5u+4=(u−1)(u−4)=0, so x²=1 or 4 and x=−2,−1,1,2. Each value satisfies the original equation.

Let u=x². Then u²−5u+4=(u−1)(u−4)=0, so x=1 or 4. Substitution gives zero for both values.

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Solve \(x^4~-~5x^2~+~4~=~0\) by rewriting it as a quadratic in \(x^2\). Give the complete real solution set and check the values in the original equation.

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