Solving Quadratic Equations

What this move does

A polynomial equation with one unknown whose highest power is $2$ always looks like this once it is tidied up:

a x^{2} + b x + c = 0 (a \neq = 0)

There are four roads to solving it. They all give exactly the same roots; they just take different amounts of effort:

Tidy up first: move every term to the left to get $a x^{2} + b x + c = 0$ , and read off $a$ , $b$ and $c$ (with their signs).
Pick the easy road first: if it can be written as $()^{2} =$ a number, just take square roots; if the left side works with the cross method, factor it into two linear factors and set each factor equal to $0$ .
Otherwise, complete the square or use the formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ . The formula is really just “complete the square once and write down the result”.
Substitute back to check, especially after moves like clearing denominators or taking square roots.

The graph above says the same thing: the roots of $a x^{2} + b x + c = 0$ are the points where the parabola $y = a x^{2} + b x + c$ lands on the $x$ -axis.

Predict first

Sam solves $x^{2} = 5 x$ : “Divide both sides by $x$ to get $x = 5$ .” How many real roots does this equation actually have?

1

, namely

5

2

0

Choose an answer first; then you can see the result.

Work through one

$x^{2} - 6 x + 4 = 0$ cannot be done with the integer cross method (there is no pair of integers with product $4$ and sum $- 6$ ). Completing the square always works.

Example: solve x² − 6x + 4 = 0 by completing the square

Step 1 / 5

Read the equation
$a = 1$ , $b = - 6$ , $c = 4$ . On the graph the parabola meets the $x$ -axis at two points, but neither is at an integer. That is why the cross method does not work.

$x^{2} - 6 x + 4 = 0$
Move the constant to the right
Add the square of half the linear coefficient to both sides
Take square roots, positive and negative
Check with the formula

Do exactly these steps for a general $a x^{2} + b x + c = 0$ and you get the quadratic formula: divide by $a$ first, then complete the square,

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}} ⟹ x = \frac{- b \pm b ^{2} - 4 a c}{2 a} .

One problem, three roads

One problem, several solutions

Solve $2 x^{2} - 5 x - 3 = 0$ .

Factoring

Cross method: $2 x^{2} - 5 x - 3 = (2 x + 1) (x - 3)$ (cross-multiply and add: $2 \cdot (- 3) + 1 \cdot 1 = - 5$ ).

If a product of two numbers is $0$ , at least one of them is $0$ : $2 x + 1 = 0$ or $x - 3 = 0$ .

So $x = - \frac{1}{2}$ or $x = 3$ .

CostFastest, but only if you can find the factors.

Formula

$a = 2, b = - 5, c = - 3$ .

$Δ = (- 5)^{2} - 4 \times 2 \times (- 3) = 25 + 24 = 49$ .

$x = \frac{5 \pm 7}{4}$ , that is, $x = 3$ or $x = - \frac{1}{2}$ .

CostNo clever guessing, but more arithmetic and easy sign slips.

Completing the square

Divide both sides by $2$ : $x^{2} - \frac{5}{2} x = \frac{3}{2}$ .

Add $(\frac{5}{4})^{2} = \frac{25}{16}$ to both sides: $(x - \frac{5}{4})^{2} = \frac{49}{16}$ .

$x - \frac{5}{4} = \pm \frac{7}{4}$ , that is, $x = 3$ or $x = - \frac{1}{2}$ .

CostThe longest, but it shows where the formula comes from.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: $a x^{2} + b x + c = 0$ can be solved with the quadratic formula

$a = 1$ , $b = 2$ , $c = - 3$ : the formula gives $x = \frac{- 2 \pm 4}{2}$ , that is, $1$ and $- 3$ , matching the two intersections on the graph.

Practice

Exercise 1Basic

Solve $(x - 2)^{2} = 9$ . (Separate the two roots with a comma.)

Exercise 2Basic

Solve $x^{2} + 4 x - 12 = 0$ .

Exercise 3Advanced

Solve $(x - 1) (x + 2) = 10$ .

Exercise 4Advanced

Complete the square to write $x^{2} - 8 x + 3 = 0$ in the form $(x - m)^{2} = n$ . Find $n$ .

Exercise 5Challenge

$x = 2$ is a root of $x^{2} + k x - 10 = 0$ . Find the other root.

What this move does

Work through one

Example: solve x² − 6x + 4 = 0 by completing the square

One problem, three roads

Factoring

Formula

Completing the square

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Dividing both sides by x and losing a root

Writing only the positive square root

Losing the sign of −b in the formula