MPM2D Real and Non-Real Roots: Notes and Practice

You throw a ball straight up from a height of 1.5 m at 14 m/s. Ignoring air resistance, its height after $t$ seconds is $h = - 4.9 t^{2} + 14 t + 1.5$ metres. Will the ball reach a shelf 10 m up? What about a ledge 12 m up? Each question is a quadratic equation: $- 4.9 t^{2} + 14 t + 1.5 = 10$ and $- 4.9 t^{2} + 14 t + 1.5 = 12$ . For the 10 m shelf you expect two answers, one on the way up and one on the way down. But if the ball never gets as high as 12 m, the second equation has no solution at all. This lesson shows how to tell which situation you are in from a single number, before you finish solving.

Investigate: three parabolas

Use graphing technology to graph $y = x^{2} - 4 x + 3$ , $y = x^{2} - 4 x + 4$ and $y = x^{2} - 4 x + 5$ . The three parabolas have the same shape and the same axis of symmetry, $x = 2$ , because only the constant term changes. Each change of 1 in the constant lifts the parabola 1 unit, so the vertices are $(2, - 1)$ , $(2, 0)$ and $(2, 1)$ . The first vertex is below the $x$ -axis, so that parabola crosses the axis twice. The second vertex is on the axis, so that parabola touches it once. The third vertex is above the axis and the parabola opens upward, away from the axis, so it never meets it.

Fig 1The parabolas y = x² − 4x + 3, y = x² − 4x + 4 and y = x² − 4x + 5, with vertices (2, −1), (2, 0) and (2, 1). The first crosses the x-axis at 1 and 3, the second touches it at 2, and the third does not reach it. Each panel shows the value of b² − 4ac.

Equation	b² − 4ac	Quadratic formula	Roots	x-intercepts
$x^{2} - 4 x + 3 = 0$	$16 - 12 = 4$	$x = \frac{4 \pm 4}{2} = \frac{4 \pm 2}{2}$	$x = 3$ or $x = 1$	2
$x^{2} - 4 x + 4 = 0$	$16 - 16 = 0$	$x = \frac{4 \pm 0}{2} = \frac{4}{2}$	$x = 2$ only	1
$x^{2} - 4 x + 5 = 0$	$16 - 20 = - 4$	$x = \frac{4 \pm - 4}{2}$	no real roots	0

The quadratic formula applied to the three equations from the investigation

The table shows what decides the case: the expression under the square root, $b^{2} - 4 a c$ . It is called the discriminant, because it discriminates (tells apart) the three cases. When it is positive, the $\pm$ adds and subtracts a real, non-zero amount, giving two different roots. When it is zero, adding and subtracting $0 = 0$ gives the same root twice, so there is one distinct root, $x = - \frac{b}{2 a}$ , which is the vertex sitting on the $x$ -axis. When it is negative, the formula asks for the square root of a negative number. No real number squares to a negative: a positive times a positive is positive, a negative times a negative is positive, and $0^{2} = 0$ . So $- 4$ is not a real number, and the equation has no real roots.

The discriminant test

For

a x^{2} + b x + c = 0

with

a \neq = 0

, calculate

b^{2} - 4 a c

. If

b^{2} - 4 a c > 0

: two distinct real roots, and the parabola crosses the

x

-axis twice. If

b^{2} - 4 a c = 0

: one real root (two equal roots), and the vertex touches the

x

-axis. If

b^{2} - 4 a c < 0

: no real roots, and the parabola does not meet the

x

-axis.

What does "no real roots" really mean? Mathematicians have built a larger number system in which negative numbers do have square roots. In that system, $x^{2} - 4 x + 5 = 0$ has two non-real roots. You will meet those numbers in later courses. In this course, when the discriminant is negative, the correct conclusion is that the equation has no real roots, and the graph of the relation has no $x$ -intercepts. Graphing technology confirms it: zoom out as far as you like, and the parabola $y = x^{2} - 4 x + 5$ stays above the $x$ -axis, because its lowest point is $(2, 1)$ .

VideoCalculate b squared minus 4ac and use its sign to tell whether a quadratic has two, one or no real roots.Watch on YouTube

Counting zeros from vertex form

If the relation is in vertex form, $y = a (x - h)^{2} + k$ , you can count its zeros without any calculation. The vertex is $(h, k)$ , so $k$ tells you whether the vertex is above, on, or below the $x$ -axis, and the sign of $a$ tells you whether the parabola opens up or down. Two zeros happen when the parabola opens toward the $x$ -axis from the vertex. For example, $y = - 2 (x - 3)^{2} + 5$ has its vertex at $(3, 5)$ , above the axis, and opens down, so it crosses the $x$ -axis twice.

Sign of a	Value of k	Where the vertex is	Number of zeros
$a > 0$ (opens up)	$k < 0$	below the x-axis	2
$a > 0$ (opens up)	$k > 0$	above the x-axis	0
$a < 0$ (opens down)	$k > 0$	above the x-axis	2
$a < 0$ (opens down)	$k < 0$	below the x-axis	0
either sign	$k = 0$	on the x-axis	1

Number of zeros of y = a(x − h) squared + k

Loading interactive…

Drag the vertex of a parabola up and down and watch its zeros, its factored form and its graph change together.

Back to the ball

For the 12 m ledge, subtract 12 to get standard form: $- 4.9 t^{2} + 14 t - 10.5 = 0$ . Here $a = - 4.9$ , $b = 14$ and $c = - 10.5$ , so $b^{2} - 4 a c = 196 - 4 (- 4.9) (- 10.5) = 196 - 205.8 = - 9.8 .$ The discriminant is negative: there are no real roots, so the ball never reaches 12 m. For the 10 m shelf, $- 4.9 t^{2} + 14 t - 8.5 = 0$ gives $b^{2} - 4 a c = 196 - 166.6 = 29.4$ , which is positive, so the ball passes 10 m twice. The formula gives $t = \frac{- 14 \pm 29.4}{- 9.8}$ , which is about $0.88$ s (going up) and $1.98$ s (coming down). The boundary case is a height of 11.5 m: $- 4.9 t^{2} + 14 t - 10 = 0$ has $b^{2} - 4 a c = 196 - 196 = 0$ , so the ball reaches 11.5 m exactly once, at the top of its path. Its maximum height is 11.5 m, reached at $t = - \frac{b}{2 a} = \frac{- 14}{- 9.8} \approx 1.43$ s.

Two more uses of the discriminant

When

a

b

and

c

are integers and

b^{2} - 4 a c

is a perfect square (0, 1, 4, 9, 16, ...), the roots are rational and the equation factors with integers; otherwise go straight to the formula. And when an equation contains an unknown constant, the discriminant finds it:

x^{2} + 6 x + k = 0

has

b^{2} - 4 a c = 36 - 4 k

, which is 0 when

k = 9

. So it has one real root when

k = 9

, two real roots when

k < 9

, and no real roots when

k > 9

Real and Non-Real Roots

Key things to remember

Investigate: three parabolas

Counting zeros from vertex form

Back to the ball

Worked examples

Where marks get lost

Flashcards 7

Practice