Foundations path · F05

Before this lesson: Algebraic identities and factorisation

Your goal: Solve equations while preserving and checking the full solution set.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Quadratic Equations and the Quadratic Formula: the key idea

An equation specifies which allowed inputs make two expressions equal. Adding the same quantity to both sides preserves the solution set. Multiplication by a nonzero number is reversible, but multiplication by an expression that can vanish needs a case check. For ax²+bx+c=0 with a≠0, completing the square gives the quadratic formula; the discriminant b²−4ac decides the number of real roots. Squaring may create extra candidates, so always return to the original equation.

A worked example

Solve x+2=x\sqrt{x+2} = x over the reals.

The square root is nonnegative, so x≥0. Squaring gives x+2=x², hence (x−2)(x+1)=0. Candidates are 2 and −1, but only 2 is nonnegative. Checking 4=2\sqrt{4}=2 confirms the sole solution.

Your turn: change one thing

Solve x(x−4)=0 and explain why dividing by x is risky.

Try this on paper before opening the explanation.

Compare your reasoning

The zero-product rule gives x=0 or x=4. Dividing by x assumes x≠0 and would erase the valid solution 0.

Pause and check

A trap to avoid: State the domain, keep exact values and explain why each step is allowed. A correct answer still needs a reason.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Derive the quadratic formula for ax²+bx+c=0, a≠0.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Multiply by 4a and rearrange to obtain (2ax+b)²=b²−4ac. Over the reals, a negative right side gives no solution. Otherwise 2ax+b=±b2−4ac2ax+b=\pm \sqrt{b^{2}-4ac}, and division by 2a gives x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Each operation is reversible when the square-root sign includes both choices.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. Solve 3x + 5 = 20.

Foundation

  1. 4
  2. 6
  3. 7
  4. 5
Hint

Subtract 5, then divide by 3.

Answer and reasoning

5. 3x = 15, so x = 5.

2. Which pair solves x + y = 7 and x − y = 1?

Foundation

  1. (5,2)
  2. (1,6)
  3. (4,3)
  4. (3,4)
Hint

Add the two equations.

Answer and reasoning

(4,3). Adding gives 2x=8, so x=4 and y=3.

3. Find all real roots of x² − 5x + 6 = 0.

Core

  1. 1 and 6
  2. 2 and 3
  3. Only 2
  4. −2 and −3
Hint

Look for two factors with product 6 and sum −5.

Answer and reasoning

2 and 3. The polynomial is (x−2)(x−3); the zero-product rule gives both roots.

4. How many distinct real roots does x² + 4x + 4 = 0 have?

Core

  1. 4
  2. 1
  3. 0
  4. 2
Hint

It is a perfect square.

Answer and reasoning

1. (x+2)²=0 only at x=−2; multiplicity two is one distinct root.

5. When solving x(x−4)=0, which step loses a solution?

Stretch

  1. Checking x=0
  2. Expanding the left side
  3. Dividing by x without a case split
  4. Using the zero-product rule
Hint

Could the proposed divisor be zero?

Answer and reasoning

Dividing by x without a case split. Dividing by x discards x=0. Separate that case before dividing.

6. After squaring x+2=x\sqrt{x+2}=x, which candidate must be rejected?

Stretch

  1. Both candidates
  2. Neither candidate
  3. −1
  4. 2
Hint

The original right side must be nonnegative.

Answer and reasoning

−1. At x=−1 the left side is 1 and the right side is −1, so it is an extraneous root.

7. How many real roots can a nonzero quadratic have?

Foundation

  1. Always two distinct
  2. At most one
  3. Infinitely many
  4. At most two
Hint

Use its degree or discriminant.

Answer and reasoning

At most two. A degree-two polynomial has at most two distinct roots.

8. Solve x²−7x+12=0.

Core

  1. x=−3 or −4
  2. x=2 or 6
  3. x=1 or 12
  4. x=3 or 4
Hint

Factor the quadratic.

Answer and reasoning

x=3 or 4. x²−7x+12=(x−3)(x−4).

9. Why check a root after squaring an equation?

Stretch

  1. All roots must be positive
  2. Squaring can introduce extra solutions
  3. Squaring always preserves sign
  4. Quadratics cannot be solved
Hint

a²=b² also allows a=−b.

Answer and reasoning

Squaring can introduce extra solutions. The squared equation can admit values that fail the original sign requirement.

Choose your next step

Continue to How to write a proof. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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