Year 10AC9M10A01

Completing the Square and the Discriminant

Expand, factorise and simplify expressions and solve equations algebraically, applying exponent laws involving products, quotients and powers of variables, and the distributive property.

Completing the square rewrites any quadratic so the turning point is visible. The discriminant tells you how many solutions exist before you solve.

Builds onYr 10 · Expanding and Factorising

Method

Completing the square

x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2

Halve $b$, square it, subtract it back.

Turning point form

y = (x-h)^2 + k \Rightarrow \text{vertex } (h, k)

Read the vertex straight off.

The discriminant

\Delta = b^2 - 4ac

What it tells you

\Delta > 0 \text{: two roots} \quad \Delta = 0 \text{: one} \quad \Delta < 0 \text{: none}

Check it before choosing a method.

Worked example

e.g.

\text{Write } y = x^2 - 6x + 5 \text{ in turning point form and find the number of roots.}

Take the coefficient of $x$ and halve it. $-6 \div 2 = -3$
Square it. $(-3)^2 = 9$
Add and subtract it inside the expression. $x^2 - 6x + 9 - 9 + 5$
Write the perfect square and tidy the rest. $(x-3)^2 - 4$
Read the turning point. $(3, -4)$
Now the discriminant. $\Delta = 36 - 20 = 16 > 0 \ \Rightarrow\ \text{two roots}$

Practice

Halve, square, subtract back. Check the discriminant before solving.

\text{Complete the square: } x^2 + 8x

Answer

(x+4)^2 - 16

\text{Turning point of } y = (x-2)^2 + 3

Answer

(2, 3)

\text{Find } \Delta \text{ for } x^2 + 4x + 4

Answer

0 \text{ — one root}

\text{Find } \Delta \text{ for } x^2 + x + 3

Answer

-11 \text{ — no real roots}

\text{How many roots has } 2x^2 - 5x + 1?

Answer

\Delta = 17 > 0 \ \Rightarrow\ \text{two}

\text{Why check } \Delta \text{ first?}

Answer

\text{It tells you whether a solution exists at all.}