Quadratics and Parabolas

In short: A quadratic is an expression like $a x^{2} + b x + c$ , and its graph is a parabola that opens up if $a > 0$ and down if $a < 0$ . Solve a quadratic equation by factorising, completing the square or the formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ .

Starting point: Years 9 and 10

A quadratic has the form $y = a x^{2} + b x + c$ with $a \neq = 0$ . Its graph is a parabola.

If $a > 0$ , the parabola opens upward (a smile) and has a minimum.
If $a < 0$ , it opens downward (a frown) and has a maximum.

Solving quadratic equations

To solve $a x^{2} + b x + c = 0$ , get zero on one side first. Then use one of three methods.

1. Factorise. If two things multiply to zero, one of them must be zero.

$x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0 \Rightarrow x = 3 or x = - 1$

2. Quadratic formula. It always works.

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

3. Completing the square. Halve the middle number, square it, add and subtract it.

$x^{2} + 6 x + 2 = (x + 3)^{2} - 9 + 2 = (x + 3)^{2} - 7$

Sketching a parabola

Find these four things:

Shape: up or down from the sign of $a$ .
y-intercept: put $x = 0$ , so it is $c$ .
x-intercepts: solve $y = 0$ .
Vertex: the turning point. Its x-value is $x = - \frac{b}{2 a}$ , which is also the axis of symmetry.

y = x² − 2x − 3: roots at −1 and 3, vertex halfway between them.

Quadratics

Form

y = ax² + bx + c
a > 0 smile, a < 0 frown

Solve

Factorise
Formula
Complete the square

Sketch

y-intercept is c
Roots from y = 0
Vertex at x = -b/2a

Discriminant

Δ = b² - 4ac
Number of roots

Year 11The discriminant

The part under the root, $Δ = b^{2} - 4 a c$ , is the discriminant. It tells you how many real roots there are without solving.

$Δ > 0$ : two different real roots (the parabola cuts the x-axis twice).
$Δ = 0$ : one repeated root (it just touches the axis).
$Δ < 0$ : no real roots (it misses the axis).

Example. For what values of $k$ does $x^{2} + 4 x + k = 0$ have two real roots? Need $16 - 4 k > 0$ , so $k < 4$ .

A quadratic that is always positive (positive definite) has $a > 0$ and $Δ < 0$ .

If $a$ contains $k$ (as in $k x^{2} - 4 x + 1 = 0$ ), remember $k \neq = 0$ , or the equation is not a quadratic.

To show an equation always has two real roots, show $Δ > 0$ for every $k$ . For $x^{2} + k x - 9 = 0$ : $Δ = k^{2} + 36$ , which is always positive because $k^{2} \geq 0$ .

Sum and product of the roots. If $α$ and $β$ are the roots of $a x^{2} + b x + c = 0$ , then $α + β = - \frac{b}{a} α β = \frac{c}{a}$ They answer questions without solving. Roots of opposite signs need $α β < 0$ , so $\frac{c}{a} < 0$ . If the roots of $x^{2} + k x - 4 = 0$ differ by 5, use $(α - β)^{2} = (α + β)^{2} - 4 α β$ : $25 = k^{2} + 16$ , so $k = \pm 3$ .

Quick summary

A quadratic $y = a x^{2} + b x + c$ graphs as a parabola. The sign of $a$ decides up or down.
Solve by factorising, the formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , or completing the square.
The vertex is at $x = - \frac{b}{2 a}$ , halfway between the roots.
The discriminant $Δ = b^{2} - 4 a c$ counts the real roots.

Quiz

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Try 3 questions now

Solve $x^{2} - 5 x + 6 = 0$ .
1. A $x = - 2$ or $x = - 3$
2. B $x = 2$ or $x = 3$
3. C $x = 1$ or $x = 6$
4. D $x = - 1$ or $x = 6$
Show answer
B: $x = 2$ or $x = 3$
$(x - 2) (x - 3) = 0$ .
Solve $x^{2} = 9 x$ .
1. A $x = 9$ only
2. B $x = 3$ or $x = - 3$
3. C $x = 0$ or $x = 9$
4. D $x = 0$ only
Show answer
C: $x = 0$ or $x = 9$
$x^{2} - 9 x = 0$ , so $x (x - 9) = 0$ . Never divide by $x$ , or you lose $x = 0$ .
The parabola $y = - 2 x^{2} + 5$ opens:
1. A upward
2. B downward
3. C to the right
4. D it is a straight line
Show answer
B: downward
$a = - 2$ is negative, so it is a frown.