B3 - Quadratic equations in unfamiliar forms | AQA A-Level Mathematics

Choose a quadratic-solving method, recognise quadratic structure inside a less familiar equation and return to the original variable. Keep exact answers and use domain restrictions to decide which candidate roots really solve the original problem.

Choosing a quadratic method

Can an equation with a fourth power have four real solutions, even though a quadratic has at most two? The equation $x^4-5x^2+4=0$ will give us all four. Its secret is that it is quadratic in a different expression.

First recall the smaller problem. In $2x^2+x=6$ , move every term to one side before solving. It becomes $2x^2+x-6=0$ , so a product equal to zero can reveal the roots.

You already know

A quadratic is $ax^2+bx+c$ , with $a\ne0$ . Factorisation, completing the square and the quadratic formula are alternative ways to solve its equation. The discriminant tells you whether its roots are real.

Factorisation is efficient when simple factors are visible. The zero-product rule says that a product of real numbers is zero only if at least one factor is zero. It is not a rule for a product equal to a nonzero number. This rule is not in the formula booklet: learn it.

Quadratic formula

For $ax^2+bx+c=0$ , with real coefficients and $a\ne0$ ,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Real roots require $b^2-4ac\ge0$ .

Not in the formula booklet: learn it.

The formula is useful when rational factors are not available. Keep brackets around negative coefficients and around the whole denominator.

Worked example 1: solve by factorisation

Question. Solve $2x^2+x=6$ .

Plan. Rearrange to zero, factorise, then set each factor equal to zero.

\begin{aligned} 2x^2+x-6&=0 && \text{rearrange} \\ (2x-3)(x+2)&=0 && \text{factorise} \\ 2x-3=0&\quad\text{or}\quad x+2=0 && \text{zero product} \\ x&=\frac32\quad\text{or}\quad x=-2 && \text{solve each factor} \end{aligned}

Why this step? Rearranging to zero is what makes the zero-product rule applicable.

Answer. $x=3/2$ or $x=-2$ .

Check. In the original equation, $2(3/2)^2+3/2=6$ and $2(-2)^2-2=6$ .

Change just the right side from $6$ to $5$ . The convenient rational factorisation disappears, so use the formula instead.

Worked example 2: exact roots from the formula

Question. Solve $2x^2+x=5$ , giving exact roots.

Plan. Rearrange to zero and substitute $a=2,b=1,c=-5$ into the quadratic formula.

\begin{aligned} 2x^2+x-5&=0 && \text{standard form} \\ x&=\frac{-1\pm\sqrt{1^2-4(2)(-5)}}{2(2)} && \text{substitute coefficients} \\ &=\frac{-1\pm\sqrt{41}}4 && \text{simplify the discriminant} \end{aligned}

Why this step? The irrational square root belongs in an exact answer; rounding it would change the roots.

Answer. $x=(-1+\sqrt{41})/4$ or $x=(-1-\sqrt{41})/4$ .

Check. If $x=(-1+s)/4$ , where $s=\pm\sqrt{41}$ , then $2x^2+x=(s^2-1)/8=5$ .

AQA permits calculator solutions for an ordinary numerical quadratic, without a compulsory written formula substitution. Exact form, a prescribed method, or a request to justify a result still has to be honoured. Displaying algebra is useful because an incorrect answer can still contain creditworthy method.

Watch out

Wrong: from $(2x-3)(x+2)=6$ , set each factor equal to zero. Right: first obtain a product equal to zero. The product in example 1 equals zero only after moving the $6$ .

Change the constant again and choose an efficient method yourself.

Completing the square

A completed square solves an equation by reducing it to one square equal to a constant.

The expansion $(x-2)^2=x^2-4x+4$ tells us that $x^2-4x=(x-2)^2-4$ . If the first terms have a common coefficient, retain it outside the whole bracket so the compensating constant is scaled correctly.

Solve from a square

For real $h,k$ ,

(x-h)^2=k

If $k>0$ , then $x=h\pm\sqrt k$ . If $k=0$ , then $x=h$ . If $k<0$ , there is no real solution.

Not in the formula booklet: learn it.

The symbol $\sqrt k$ is non-negative, but an equation asking which numbers square to a positive $k$ has both signs. This is the same repeated-root boundary seen on quadratic graphs.

Worked example 3: solve using a positive square

Question. Solve $2x^2-8x+3=0$ by completing the square, giving exact roots.

Plan. Form $(x-2)^2$ , isolate it, then take both square-root signs.

\begin{aligned} 2x^2-8x+3&=2(x^2-4x)+3 && \text{factor }2 \\ &=2\bigl((x-2)^2-4\bigr)+3 && \text{complete the square} \\ &=2(x-2)^2-5 && \text{simplify} \\ 2(x-2)^2-5=0&\Rightarrow (x-2)^2=\frac52 && \text{isolate the square} \\ x&=2\pm\frac{\sqrt{10}}2 && \text{both signs} \end{aligned}

Why this step? The $-4$ inside the bracket is multiplied by $2$ ; overlooking this changes both roots.

Answer. $x=2+\sqrt{10}/2$ or $x=2-\sqrt{10}/2$ .

Check. For either root, $(x-2)^2=10/4=5/2$ , so $2(x-2)^2-5=0$ .

Change only the constant from $3$ to $8$ . The square will now equal zero, so the two signs coincide.

Worked example 4: a repeated root from the square

Question. Solve $2x^2-8x+8=0$ by completing the square.

Plan. Complete the same square and find where it vanishes.

\begin{aligned} 2x^2-8x+8&=2\bigl((x-2)^2-4\bigr)+8 && \text{complete the square} \\ &=2(x-2)^2 && \text{constants cancel} \\ 2(x-2)^2=0&\Rightarrow x-2=0 && \text{square equals zero} \\ x&=2 && \text{solve} \end{aligned}

Why this step? There is only one distinct number whose square is zero, so listing $2$ twice adds no solution.

Answer. $x=2$ , a repeated root.

Check. $2(2)^2-8(2)+8=0$ .

Completed-square form and the formula must agree: both are rearrangements of the same equation. Completing the square is particularly revealing when the right side is zero or negative.

Watch out

Wrong: $(x-2)^2=5/2$ gives only $x=2+\sqrt{10}/2$ . Right: both signs are needed because both positive and negative bracket values have that square.

Move the constant past the repeated-root case and use the square to explain the result.

A quadratic in another expression

A substitution names a repeated expression, turning an unfamiliar equation into a familiar quadratic.

If $u=x+1$ , then $(x+1)^2=u^2$ . Both occurrences must change together. After finding $u$ , return to $x$ : the question has not asked for the new variable.

Quadratic in a function of the unknown

An equation of the form $af(x)^2+bf(x)+c=0$ , with $a\ne0$ , is quadratic in $f(x)$ . The substitution $u=f(x)$ gives $au^2+bu+c=0$ . For example, $x^4-5x^2+4=0$ is quadratic in $x^2$ , because $x^4=(x^2)^2$ .

The substitution's range matters. A bracket such as $x+1$ can take any real value, whereas $x^2\ge0$ . This substitution process is not in the formula booklet: learn to recognise the repeated expression.

Worked example 5: a repeated bracket

Question. Solve $(x+1)^2-5(x+1)+4=0$ .

Plan. Put $u=x+1$ , solve the quadratic, then undo the substitution.

\begin{aligned} u^2-5u+4&=0 && u=x+1 \\ (u-1)(u-4)&=0 && \text{factorise} \\ u&=1\text{ or }4 && \text{zero product} \\ x+1&=1\text{ or }4 && \text{return to }x \\ x&=0\text{ or }3 && \text{subtract }1 \end{aligned}

Why this step? Keeping the bracket intact avoids expanding away the structure you can already solve.

Answer. $x=0$ or $x=3$ .

Check. The original bracket is respectively $1$ and $4$ ; $1-5+4=0$ and $16-20+4=0$ .

Now change the repeated expression from a linear bracket to $x^2$ . Undoing this substitution can produce two original-variable roots from each positive substituted root.

Worked example 6: a fourth power hiding a quadratic

Question. Find all real solutions of $x^4-5x^2+4=0$ .

Plan. Put $u=x^2\ge0$ , solve for $u$ , then take both signs for each positive square.

\begin{aligned} u^2-5u+4&=0 && u=x^2 \\ (u-1)(u-4)&=0 && \text{factorise} \\ u&=1\text{ or }4 && \text{both non-negative} \\ x^2&=1\text{ or }4 && \text{undo the substitution} \\ x&=\pm1\text{ or }\pm2 && \text{both signs each time} \end{aligned}

Why this step? Two roots in $u$ need not mean two roots in $x$ : the substitution $u=x^2$ combines opposite values of $x$ .

Answer. $x=-2,-1,1,2$ .

Check. The original expression is zero at $x^2=1$ and $x^2=4$ , so each positive and negative pair works.

This answers the opening puzzle. There is another small surprise: a quadratic in $x^2$ can have three distinct real solutions. For $x^4-x^2=0$ , the substituted roots are $0,1$ ; the first gives only $x=0$ , while the second gives $x=\pm1$ . Count original values, not signs written on paper.

Watch out

Wrong: “ $u=1,4$ , so the solutions are $x=1,4$ .” Right: solve $x^2=1$ and $x^2=4$ . Always write the equation that undoes your substitution.

Later, the same structure will appear with trigonometric and exponential expressions; their ranges will decide which substituted roots survive. Here the square substitution is enough.

Domains and admissible roots

An algebraic candidate becomes a solution only when it satisfies the original equation and its domain.

For $u=\sqrt x$ , record $x\ge0$ and $u\ge0$ . A negative $u$ cannot be a principal square root, even if it solves the transformed quadratic. For $u=1/x$ , record $x\ne0$ and $u\ne0$ . These domain facts are not in the formula booklet: apply the definitions.

Worked example 7: reject a negative square root

Question. Solve $x-\sqrt x-6=0$ over the real numbers.

Plan. Put $u=\sqrt x\ge0$ , use $x=u^2$ , and reject any negative value of $u$ .

\begin{aligned} u^2-u-6&=0 && u=\sqrt x\ge0 \\ (u-3)(u+2)&=0 && \text{factorise} \\ u&=3\text{ or }-2 && \text{candidate values} \\ u&=3 && \text{reject }-2<0 \\ x&=3^2=9 && \text{recover }x \end{aligned}

Why this step? Squaring $u=-2$ would produce $x=4$ , but $\sqrt4=2$ , not $-2$ ; it would not undo the substitution.

Answer. $x=9$ .

Check. $9-\sqrt9-6=9-3-6=0$ ; the rejected candidate $x=4$ gives $4-2-6=-4$ .

Change just the sign before the square root. The admissible substituted root changes too; the domain rule does not.

Worked example 8: the same domain with a changed sign

Question. Solve $x+\sqrt x-6=0$ over the real numbers.

Plan. Use $u=\sqrt x\ge0$ , solve the new quadratic and return to $x$ .

\begin{aligned} u^2+u-6&=0 && u=\sqrt x\ge0 \\ (u+3)(u-2)&=0 && \text{factorise} \\ u&=-3\text{ or }2 && \text{candidate values} \\ u&=2 && \text{reject }-3<0 \\ x&=2^2=4 && \text{recover }x \end{aligned}

Why this step? The sign condition applies to the defined variable $u$ , independently of the signs in its quadratic equation.

Answer. $x=4$ .

Check. $4+2-6=0$ . Squaring the rejected $-3$ would give $x=9$ , which instead gives $9+3-6=6$ .

For a reciprocal example, $2/x^2+1/x-1=0$ with $x\ne0$ becomes $2u^2+u-1=0$ , where $u=1/x\ne0$ . The factors $(2u-1)(u+1)=0$ give $u=1/2,-1$ , so $x=2,-1$ . Both satisfy the nonzero condition. Multiplying by $x^2$ is an equally valid route on that domain.

Squaring needs an additional check because it can lose sign information. From $\sqrt{x+2}=x$ , squaring gives $x^2-x-2=0$ , with candidates $2,-1$ . The original right side must be non-negative; $-1$ fails, while $2$ works. Squaring preserved a true solution but also introduced another candidate.

Watch out

Wrong: solve the transformed quadratic and accept every root. Right: apply the substitution's range, undo it, then check the original equation. Multiplication by a known nonzero quantity is reversible on its domain; squaring does not preserve sign information.

In the next question only the constant changes. Keep the non-negative square-root condition when solving the new quadratic.

Quick Recap

Rearrange a quadratic to zero before using the zero-product rule; choose factorisation, completing the square or the formula efficiently.
A positive square gives both signs; a zero square gives one distinct root; a negative square gives no real root.
Name the repeated expression, record its possible values and undo the substitution after solving.
Check admissibility and the original equation, especially after squaring or clearing denominators.

The opening fourth-power equation has four real solutions because each positive root of its quadratic in $x^2$ gives two opposite values of $x$ : $-2,-1,1,2$ .

Last Accessed

Pure Mathematics

Mechanics

Statistics

B3 - Quadratic equations in unfamiliar forms

Choosing a quadratic method

Completing the square

A quadratic in another expression

Domains and admissible roots

Quick Recap

Learn faster with Nova