GCSE Maths · Algebra

Quadratic Equations flashcards

12 cards on Quadratic Equations: the key points, worked examples and the slips to watch. Swipe through them, or print them.

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Quadratic Equations

Solving by factorising

How do you solve a quadratic equation such as x² + 3x = 10 by factorising?

Answer it in your head, then turn the card over.

Solving by factorising

Key points

Rearrange so one side is 0: x² + 3x − 10 = 0.
Factorise: find two numbers that multiply to the number term and add to the x coefficient.
If two brackets multiply to 0, one of them is 0: set each bracket equal to 0 and solve.

x² + 3x − 10 = 0

Multiply to −10, add to 3: the numbers are 5 and −2

(x + 5)(x − 2) = 0

x + 5 = 0 or x − 2 = 0, so x = −5 or x = 2

Watch out

Factorising before rearranging: x(x + 3) = 10 does not mean either factor is 10.

Quadratic Equations

Reading the solutions from the brackets

Once a quadratic is factorised, how do you get the solutions from the brackets?

Answer it in your head, then turn the card over.

Reading the solutions from the brackets

Key points

Set each bracket equal to 0 and solve it.
So each solution has the opposite sign to the number in its bracket.
Check by putting each solution back into the original equation.

x² + 5x + 6 = 0

(x + 2)(x + 3) = 0

x = −2 or x = −3

Check x = −2: 4 − 10 + 6 = 0

Watch out

Copying the signs from the brackets: (x + 2)(x + 3) = 0 gives −2 and −3, not 2 and 3.

Quadratic Equations

No number term, or no x term

How do you solve quadratics such as x² − 5x = 0 and x² − 9 = 0?

Answer it in your head, then turn the card over.

No number term, or no x term

Key points

No number term: take out x as a common factor, so one solution is always x = 0.
No x term, as in x² − 9: it is a difference of two squares, (x + 3)(x − 3).
Or rearrange to x² = 9 and square-root, keeping both signs: x = ±3.

Example

x² − 5x = 0: x(x − 5) = 0, so x = 0 or x = 5

x² − 9 = 0: (x + 3)(x − 3) = 0, so x = 3 or x = −3

Watch out

Dividing both sides of x² − 5x = 0 by x loses the solution x = 0.

Quadratic Equations

Roots on a graph

How do you read the solutions of a quadratic equation from its graph?

Answer it in your head, then turn the card over.

Roots on a graph

Key points

The roots are where the curve crosses the x-axis: the x values that make y = 0.
Two crossings means two roots; touching the axis once means one repeated root; missing it means no real roots.
The turning point is halfway between the two roots, on the line of symmetry.

A curve crossing at x = 1 and x = 5

The roots are x = 1 and x = 5

Line of symmetry: x = (1 + 5) ÷ 2 = 3

Watch out

Read where the curve crosses the x-axis, not the y-axis: that crossing is the number term.

Quadratic Equations

Drawing a quadratic graph

How do you make a table of values and draw the graph of a quadratic?

Answer it in your head, then turn the card over.

Drawing a quadratic graph

Key points

Substitute each x value into the equation, with brackets around negatives.
Plot the points and join them with a smooth U-shaped curve, not straight lines.
The y values mirror each other either side of the line of symmetry: a good check.

y = x² − 2x − 3

x = −1: 1 + 2 − 3 = 0

x = 1: 1 − 2 − 3 = −4, the lowest point

x = 3: 9 − 6 − 3 = 0, so the roots are −1 and 3

Watch out

Typing −3² on a calculator gives −9: type (−3)², which is 9.

Quadratic Equations

Higher

When x² has a number in front

How do you factorise and solve a quadratic such as 2x² + 7x + 3 = 0?

Answer it in your head, then turn the card over.

When x² has a number in front

Key points

Multiply the x² number by the number term: 2 × 3 = 6.
Find two numbers that multiply to that and add to the x coefficient, then split the middle term.
Factorise in pairs, then set each bracket equal to 0.
A root is often a fraction: 2x + 1 = 0 gives x = $\frac{- 1}{2}$ .

2x² + 7x + 3 = 0

Multiply to 6, add to 7: 6 and 1

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3)

(2x + 1)(x + 3) = 0, so x = $\frac{- 1}{2}$ or x = −3

Watch out

(2x + 1) = 0 gives x = $\frac{- 1}{2}$ , not x = −1: you still have to divide by 2.

Quadratic Equations

Higher

The quadratic formula

How do you solve a quadratic equation with the quadratic formula?

Answer it in your head, then turn the card over.

The quadratic formula

Key points

Rearrange to ax² + bx + c = 0 and write down a, b and c, with their signs.
Put negative values in brackets before you substitute.
The ± gives two solutions: one adding the root, one taking it away.
Use it when the quadratic will not factorise; round only at the very end.

Formula

x = $(- b \pm \sqrt{b^{2} - 4 a c})$ ÷ 2a

x² − 4x + 1 = 0

a = 1, b = −4, c = 1

b² − 4ac = (−4)² − 4 × 1 × 1 = 12

x = $(4 \pm \sqrt{12})$ ÷ 2 = 3.73 or 0.27 (2 dp)

Watch out

Without brackets, −4² becomes −16: it is (−4)² = 16.

Quadratic Equations

Higher

How many solutions

How does b² − 4ac tell you how many solutions a quadratic equation has?

Answer it in your head, then turn the card over.

How many solutions

Key points

b² − 4ac is the part under the square root in the formula.
Positive: two different solutions, and the graph crosses the x-axis twice.
Zero: one repeated solution, and the graph just touches the x-axis.
Negative: no real solutions, because no real number squares to a negative.

Example

x² + 2x + 5 = 0: 2² − 4 × 1 × 5 = −16, so no real solutions

x² − 6x + 9 = 0: (−6)² − 4 × 1 × 9 = 0, so one solution

Watch out

A negative b² − 4ac does not mean negative solutions: it means there are none.

Quadratic Equations

Higher

Completing the square

How do you write x² + bx + c in completed-square form?

Answer it in your head, then turn the card over.

Quadratic Equations

Higher

Solving by completing the square

How do you use completed-square form to solve a quadratic equation exactly?

Answer it in your head, then turn the card over.

Solving by completing the square

Key points

Write the quadratic in completed-square form and set it equal to 0.
Move the number to the other side, then square-root both sides, keeping ±.
Exact answers leave the root as a surd, such as $\sqrt{7}$ .

x² + 6x + 2 = 0

(x + 3)² − 7 = 0, so (x + 3)² = 7

x + 3 = $\pm \sqrt{7}$ , so x = −3 ± $\sqrt{7}$

As decimals: x = −0.35 or x = −5.65 (2 dp)

Watch out

Square-rooting gives two answers: x + 3 = $\sqrt{7}$ alone loses x = −3 − $\sqrt{7}$ .

Quadratic Equations

Higher

Turning points

How does completed-square form give you the turning point of a quadratic graph?

Answer it in your head, then turn the card over.

Turning points

Key points

y = (x + p)² + q has its turning point at (−p, q).
The bracket squared is never negative and is 0 when x = −p, which leaves y = q.
For x² + bx + c, that is a minimum: the curve is U-shaped.

y = x² + 8x + 5

8 ÷ 2 = 4: y = (x + 4)² − 16 + 5

y = (x + 4)² − 11

Turning point: (−4, −11)

Watch out

The x-coordinate has the opposite sign to the bracket: (x − 2)² + 7 turns at (2, 7).

Quadratic Equations

Quadratics from worded problems

How do you set up and solve a quadratic equation from an area problem?

Answer it in your head, then turn the card over.

Quadratics from worded problems

Key points

Write the area as a product: a rectangle x by (x + 3) has area x(x + 3) = x² + 3x.
Set it equal to the area given, then rearrange to = 0.
Solve, then check each root against the story: a length cannot be negative.

Rectangle x by (x + 3), area 40

x² + 3x = 40, so x² + 3x − 40 = 0

(x + 8)(x − 5) = 0, so x = −8 or x = 5

A length cannot be −8, so x = 5: the sides are 5 and 8

Watch out

Give a reason when you reject a root: say why −8 cannot be a length.

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