Why does solving x squared equals 5x give x = 0 or x = 5, not just x = 5?

Because dividing both sides by x assumes x is not zero, but x = 0 is one of the solutions. Instead, rearrange to x squared minus 5x equals 0, factorise to x times (x minus 5) equals 0, and use the null-factor law: a product is zero only when a factor is zero. So x = 0 or x minus 5 = 0, giving x = 0 or x = 5. Dividing by x deletes the x = 0 root.

Why must a quadratic be set equal to zero before factorising?

Because the null-factor law only works against zero: a times b equals 0 means a = 0 or b = 0. Against any other number the law fails, since 2 times 3 equals 6 with neither factor equal to 6. So you cannot set each bracket equal to a non-zero right-hand side. Collect everything to one side so the equation reads = 0, then factorise.

How do you solve a quadratic set in a context such as a length?

Solve the quadratic fully to find both roots, then reject any root the context makes impossible. For x squared plus x minus 90 equals 0, factorise to (x plus 10)(x minus 9) equals 0, giving x = minus 10 or x = 9. If x is a length, a negative value is impossible, so reject x = minus 10 and keep x = 9.

GCSE Maths Higher · AQA · Quadratics: solving & reading roots

Solving a quadratic: rearrange to = 0 before factorising

Faced with $x^2 = 5x$ , the instinct is to divide both sides by $x$ and write $x = 5$ . That answer is incomplete: it silently deletes the root $x = 0$ . You cannot divide by $x$ , because $x$ might be zero, and dividing by zero is not allowed. The reliable method is to move everything to one side so the equation reads $= 0$ : $x^2 - 5x = 0$ , factorise $x(x - 5) = 0$ , and read both roots: $x = 0 \text{ or } x = 5$ .

The principle is the null-factor law: a product equals zero exactly when at least one factor is zero. It only works against zero, which is why you must rearrange first and why you cannot set each bracket equal to a non-zero side. Dropping a root this way is one of the most reliable grade 7 to 9 mark-losers on AQA Higher papers.

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How to spot it in your own work

You divided $x^2 = 5x$ by $x$ to get only $x = 5$ , losing the root $x = 0$ .
You set each bracket of $(x + 2)(x - 5) = 6x$ equal to $6x$ instead of collecting to $x^2 - 9x - 10 = 0$ first.
You read $(x + 2)(x - 5) = 6x$ as $x = -2 \text{ or } x = 5$ by setting each bracket to zero without subtracting the $6x$ .
You kept an impossible root in a context: writing $x = -10$ for a length when $x^2 + x - 90 = 0$ gives $x = -10 \text{ or } x = 9$ and only $x = 9$ is valid.

An exam question that triggers it

Here is the structure of a typical AQA Higher quadratic question:

Solve $x^2 = 5x$ .

Write down all the values of $x$ .

The misconception divides by $x$ to get $x = 5$ only. The fix: rearrange to $x^2 - 5x = 0$ , factorise $x(x - 5) = 0$ , and read a root from each factor: $x = 0 \text{ or } x = 5$ .

Why students fall for this

Dividing by $x$ feels efficient: it strips the equation down to something linear-looking in one step. But it assumes $x \neq 0$ , and that assumption throws away the root $x = 0$ whenever it is a solution. The lost root is invisible at the moment the division is done, so the slip goes unnoticed.

A second pattern appears with brackets. For $(x + 2)(x - 5) = 6x$ students set each bracket equal to the right-hand side, writing $x + 2 = 6x$ and $x - 5 = 6x$ . This misuses the null-factor law, which only works against zero. Since $2 \times 3 = 6$ with neither factor equal to 6, the law gives no information against a non-zero number.

The mirror-image error is over-applying the rule. A genuinely linear equation such as $3x + 7 = 2x$ has no $x^2$ term, so there is nothing to set to zero and factorise: isolating $x$ to get $x = -7$ is exactly right. The signal is whether an $x^2$ term is present.

The fix: Rearrange to = 0, factorise, then read a root from each factor

Step 1: check for an x² term. If there is one, it is a quadratic: continue below. If there is no $x^2$ term it is linear, so just isolate $x$ .

Step 2: collect everything to one side. For $x^2 = 5x$ , subtract $5x$ from both sides: $x^2 - 5x = 0$ . Never divide by $x$ .

Step 3: factorise the non-zero side. $x^2 - 5x = x(x - 5)$ , so $x(x - 5) = 0$ .

Step 4: apply the null-factor law. A product is zero only when a factor is zero: $x = 0 \text{ or } x - 5 = 0$ , giving $x = 0 \text{ or } x = 5$ . Read one root from each factor.

Step 5: reject impossible roots in context. If $x$ is a length or a count, discard any negative or otherwise impossible root, keeping the valid value.

Worked example

Solve $x^2 = 5x$ .

Collect to = 0. $x^2 - 5x = 0$ . Do not divide by $x$ .
Factorise. $x(x - 5) = 0$ .
Read both roots.
$x = 0 \quad \text{or} \quad x = 5$
Dividing by $x$ would give only $x = 5$ , dropping $x = 0$ .

Solve $(x + 2)(x - 5) = 6x$ .

Expand the left. $(x + 2)(x - 5) = x^2 - 3x - 10$ .
Collect to = 0. $x^2 - 3x - 10 = 6x \Rightarrow x^2 - 9x - 10 = 0$ .
Factorise and read both roots.
$(x - 10)(x + 1) = 0 \Rightarrow x = 10 \;\text{or}\; x = -1$
Trap: setting each bracket to zero without collecting gives $x = -2 \text{ or } x = 5$ , wrong.
Verify. At $x = 10$ : $(12)(5) = 60 = 6 \times 10$ ✓. At $x = -1$ : $(1)(-6) = -6 = 6 \times (-1)$ ✓.

A rectangle's dimensions satisfy $x^2 + x - 90 = 0$ , where $x$ is a length in cm. Find $x$ .

Factorise. $(x + 10)(x - 9) = 0$ (factors of $-90$ summing to $+1$ are $+10$ and $-9$ ).
Read both roots. $x = -10 \text{ or } x = 9$ .
Reject the impossible root. A length cannot be negative, so discard $x = -10$ :
$x = 9 \text{ cm}$
Check: $9^2 + 9 - 90 = 81 + 9 - 90 = 0$ ✓.

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Common questions

Why is the answer to $x^2 = 5x$ $x = 0 \text{ or } x = 5$ and not just $x = 5$ ?: Because dividing by $x$ assumes $x \neq 0$ , but $x = 0$ is a solution. Rearrange to $x^2 - 5x = 0$ , factorise $x(x - 5) = 0$ , and by the null-factor law $x = 0 \text{ or } x = 5$ . Dividing by $x$ deletes the $x = 0$ root.
Can I set each bracket of $(x + 2)(x - 5) = 6x$ equal to $6x$ ?: No. The null-factor law only works against zero, because $2 \times 3 = 6$ with neither factor equal to 6. Expand to $x^2 - 3x - 10$ , collect to $x^2 - 9x - 10 = 0$ , then factorise $(x - 10)(x + 1) = 0$ for $x = 10 \text{ or } x = -1$ .
How do I solve a quadratic when $x$ is a length or a count?: Solve fully to find both roots, then reject any root the context makes impossible. For $x^2 + x - 90 = 0$ with $x$ a length, $(x + 10)(x - 9) = 0$ gives $x = -10 \text{ or } x = 9$ ; a length cannot be negative, so the answer is $x = 9$ .

Related misconceptions

Completing the square: forgetting to subtract the adjustment constantWriting (x+4)²−5 for x²+8x−5 instead of (x+4)²−21, keeping the original constant rather than subtracting the hidden a², a companion Higher quadratics misconception.
Reading the roots of a quadratic from the wrong graph featureTaking the roots from the turning point or the y-intercept instead of the x-axis crossings, a sibling Higher quadratics misconception about reading solutions correctly.

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