Why dividing by x loses a root in x² − 5x = 0

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Question

Solve x² − 5x = 0. A student divides both sides by x and obtains only x = 5. Which root was lost, and why?

Why dividing by 'x' in x²-5x=0 loses a rootWatch on YouTube

We need to solve the equation without performing an operation that removes a possible case. We will factor first, solve each factor separately, check every candidate in the original equation and only then state the complete solution set.

Step 1: factor out the common variable

Both terms in the quadratic equation x25x=0x^2 - 5x = 0 share a common variable factor xx. Decompose each term and factor xx outside the brackets:

x25x=0(Given quadratic equation)x^2 - 5x = 0 \quad (\text{Given quadratic equation})
xx5x=0(Identify common factor x)x \cdot x - 5 \cdot x = 0 \quad (\text{Identify common factor } x)
x(x5)=0(Factor out x)x(x - 5) = 0 \quad (\text{Factor out } x)

Step 2: apply the zero-product rule

By the Zero-Product Property, if the product of two real expressions equals zero (AB=0A \cdot B = 0), then at least one factor must equal zero (A=0A = 0 or B=0B = 0).

x(x5)=0x(x - 5) = 0
Applying Zero-Product Property: x=0orx5=0\text{Applying Zero-Product Property: } x = 0 \quad\text{or}\quad x - 5 = 0
Case 1: x=0\text{Case 1: } x = 0
Case 2: x5=0    x5+5=0+5    x=5\text{Case 2: } x - 5 = 0 \implies x - 5 + 5 = 0 + 5 \implies x = 5
Combined solution set: x=0orx=5\text{Combined solution set: } x = 0 \quad\text{or}\quad x = 5

Step 3: check both candidates in the original equation

Check the first candidate (x=0x = 0):

LHS=(0)25(0)\text{LHS} = (0)^2 - 5(0)
LHS=00=0\text{LHS} = 0 - 0 = 0
LHS=RHS=0(Valid root )\text{LHS} = \text{RHS} = 0 \quad (\text{Valid root } \checkmark)

Check the second candidate (x=5x = 5):

LHS=(5)25(5)\text{LHS} = (5)^2 - 5(5)
LHS=2525=0\text{LHS} = 25 - 25 = 0
LHS=RHS=0(Valid root )\text{LHS} = \text{RHS} = 0 \quad (\text{Valid root } \checkmark)

Both substitutions evaluate to true identity statements (0=00 = 0), confirming that the equation has two distinct real roots.

Step 4: identify what division by x assumes

Here is the incorrect method that students frequently attempt:

x25x=0x^2 - 5x = 0
Dividing both sides by x implicitly assumes x0\text{Dividing both sides by } x \text{ implicitly assumes } x \neq 0
x25xx=0x\frac{x^2 - 5x}{x} = \frac{0}{x}
x2x5xx=0\frac{x^2}{x} - \frac{5x}{x} = 0
x5=0    x=5x - 5 = 0 \implies x = 5
Warning: The root x=0 was discarded because dividing by zero is undefined!\text{Warning: The root } x = 0 \text{ was discarded because dividing by zero is undefined!}

The reusable rule

  • If every term contains a variable factor, factor it out first.
  • Never cancel a variable that could equal zero unless the zero case has already been considered.
  • Check the number of roots. A quadratic can have two, one or no real roots; an operation must not change its valid solution set.
  • Substitute answers into the original equation, not only into a later equation that may have lost a case.

Final step: state every verified root

x=0orx=5\boxed{x = 0 \quad\text{or}\quad x = 5}

Young Math Brains asks what a cancellation assumes before allowing it, then verifies every root in the original quadratic equation.

Practise factorisation without losing roots

Your turn

What assumption do you make automatically when you divide both sides of an equation by x?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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