1.1 - Equations in quadratic form | Cambridge A-Level Mathematics

An equation can hide a quadratic inside powers, square roots or reciprocals. You will identify a useful replacement variable, solve the resulting quadratic and return carefully to the original input. Conditions on the replacement variable will explain why some algebraic candidates must be rejected.

Choosing a new variable

How can $x^4-13x^2+36=0$ be solved without learning a new formula for fourth powers? Notice that $x^4=(x^2)^2$ : the equation is quadratic in $x^2$ .

Set $u=x^2$ . The equation becomes $u^2-13u+36=0$ , which uses the methods you already know. The job has two stages: solve for $u$ , then solve $x^2=u$ . Because $x$ is real, remember from the start that $u\ge0$ .

Quadratic in an expression

An equation is quadratic in an expression when it can be written $au^2+bu+c=0$ , with $a\ne0$ , after replacing that repeated expression by $u$ . A biquadratic equation such as $ax^4+bx^2+c=0$ is quadratic in $x^2$ .

Choose the repeated expression, not its square: replacing $x^4$ by $u$ would leave an awkward $x^2$ term rather than a quadratic in $u$ .

Worked example 1: four real answers

Question. Solve $x^4-13x^2+36=0$ for real $x$ .

Plan. Put $u=x^2$ , solve the quadratic in $u$ , then take both signs for each positive value.

\begin{aligned} u^2-13u+36&=0\\ (u-4)(u-9)&=0\\ u&=4\text{ or }9 \end{aligned}

Thus $x^2=4$ gives $x=\pm2$ , and $x^2=9$ gives $x=\pm3$ .

Answer. $x=-3,-2,2,3$ .

Check. At $x=\pm2$ , the original is $16-52+36=0$ . At $x=\pm3$ , it is $81-117+36=0$ . Only even powers occur, so each sign pair gives the same value.

Keep the first two terms but change the constant. One of the new values of $u$ will violate $u\ge0$ .

Worked example 2: an impossible value of the square

Question. Solve $x^4-13x^2-14=0$ for real $x$ .

Plan. Use the same replacement and check each transformed root before returning to $x$ .

\begin{aligned} u^2-13u-14&=0\\ (u-14)(u+1)&=0\\ u&=14\text{ or }-1 \end{aligned}

Since $u=x^2\ge0$ , reject $u=-1$ . From $x^2=14$ , take both signs.

Answer. $x=\pm\sqrt{14}$ .

Check. Both proposed roots have $x^2=14$ and $x^4=196$ , giving $196-182-14=0$ . No real number has square $-1$ .

A fourth-degree equation can therefore have fewer than four distinct real roots. The transformed equation alone does not decide which values return to real inputs.

Watch out

Stopping at $u=4,9$ has not solved an equation in $x$ . Write the substitution explicitly, then return to $x$ ; remember both signs when solving $x^2=$ a positive number.

Use a new quadratic in $x^2$ .

Returning to the original variable

The substitution tells you how to return to the original variable. If $u=\sqrt x$ , then $x=u^2$ ; if $u=1/x$ , then $x=1/u$ . These are different return steps even if the quadratic in $u$ looks identical.

Write the restrictions alongside the substitution. For $u=\sqrt x$ , the original requires $x\ge0$ , and the square-root symbol requires $u\ge0$ . For $u=1/x$ , $x\ne0$ and $u\ne0$ . A transformed root outside these restrictions cannot represent an original input.

Worked example 3: replacing a square root

Question. Solve $2x-3\sqrt x+1=0$ .

Plan. Put $u=\sqrt x\ge0$ , so $x=u^2$ , and solve the resulting quadratic.

\begin{aligned} 2u^2-3u+1&=0\\ (2u-1)(u-1)&=0\\ u&=\frac12\text{ or }1 \end{aligned}

Both roots satisfy $u\ge0$ . Squaring each gives the original input.

Answer. $x=1/4$ or $x=1$ .

Check. At $x=1/4$ , $2x-3\sqrt x+1=1/2-3/2+1=0$ . At $x=1$ , it is $2-3+1=0$ .

Now use the same quadratic coefficients but replace the repeated expression by a reciprocal. The return step changes from squaring to taking a reciprocal.

Worked example 4: replacing a reciprocal

Question. Solve $2/x^2-3/x+1=0$ .

Plan. Put $u=1/x\ne0$ , so $1/x^2=u^2$ , and solve for $u$ .

\begin{aligned} 2u^2-3u+1&=0\\ (2u-1)(u-1)&=0\\ u&=\frac12\text{ or }1 \end{aligned}

Both values are nonzero. Since $x=1/u$ , they give $x=2$ and $x=1$ , respectively.

Answer. $x=1$ or $x=2$ .

Check. At $x=2$ , the original is $2/4-3/2+1=0$ ; at $x=1$ , it is $2-3+1=0$ . Neither answer makes a denominator zero.

In both examples the intermediate roots were $1/2,1$ . The original answers differed because the substitutions differed. This is why the substitution must remain visible throughout the solution.

Watch out

If $u=\sqrt x$ , do not write $x=\pm u^2$ : $x$ is simply $u^2$ . The two-sign step belongs to solving $x^2=u$ , which is a different substitution.

Use a reciprocal substitution in this check.

Rejecting impossible values

Solving the transformed quadratic gives candidates, rather than an automatic list of original solutions. Check the substitution's restrictions before reversing it, then substitute the remaining answers into the original equation.

In particular, $u=\sqrt x$ cannot be negative. Squaring a negative candidate would produce a non-negative $x$ , but $\sqrt{x}$ would then be its positive magnitude, not that negative candidate. The original equation exposes the mistake.

Worked example 5: rejecting a negative square-root candidate

Question. Solve $x+\sqrt x-6=0$ .

Plan. Use $u=\sqrt x\ge0$ , solve for $u$ , and reject any negative candidate before squaring.

\begin{aligned} u^2+u-6&=0\\ (u+3)(u-2)&=0\\ u&=-3\text{ or }2 \end{aligned}

Reject $u=-3$ , because $u=\sqrt x\ge0$ . The valid candidate $u=2$ gives $x=4$ .

Answer. $x=4$ only.

Check. $4+\sqrt4-6=0$ . Squaring the rejected value would give $x=9$ , but $9+\sqrt9-6=6\ne0$ : it is not a solution.

Change only the sign before the square root. The quadratic's linear coefficient changes, so the accepted and rejected magnitudes exchange roles.

Worked example 6: a different valid candidate

Question. Solve $x-\sqrt x-6=0$ .

Plan. Use the same non-negative substitution and test the new roots against it.

\begin{aligned} u^2-u-6&=0\\ (u-3)(u+2)&=0\\ u&=3\text{ or }-2 \end{aligned}

Reject $u=-2$ ; $u=3$ gives $x=9$ .

Answer. $x=9$ only.

Check. $9-\sqrt9-6=0$ . The rejected candidate would give $x=4$ , but $4-\sqrt4-6=-4\ne0$ .

The original equation is the final authority. A valid return step plus substitution into the original checks that no denominator is zero, no square-root sign has changed and the stated variable is the one you have actually solved for. The same habit will be useful when trigonometric expressions become the replacement variable later.

Problem-solving

Write three lines of decisions: define the substitution and its restrictions; solve for the new variable; return to the original variable and check. Do not square a forbidden value merely to obtain a real-looking answer.

The opening fourth-power equation needed only a quadratic in $x^2$ . Its four real answers came from two permitted positive values of that square, each giving two signs.

Quick Recap

A repeated expression can hide a quadratic; choose it as a new variable.
State its restrictions, such as $x^2\ge0$ , $\sqrt x\ge0$ or $1/x\ne0$ .
Reverse the actual substitution: square, take square roots or take reciprocals as appropriate.
Check the original equation after rejecting impossible transformed values.

The opening equation has $x^2=4$ or $9$ , so its real roots are $-3,-2,2,3$ .

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1.1 - Equations in quadratic form

Choosing a new variable

Returning to the original variable

Rejecting impossible values

Quick Recap

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