Quadratic Functions · Little Blue Book, Visualized

Try this

Move the parabola up (increase $c$ ) until the two roots merge into one, then keep going. Which condition turns ✗ first?
Move only $b$ and push the axis out of the interval. The values at both ends may still be positive, but where have the actual roots gone?
Switch to “m between the roots”. Why is one condition enough now?
Switch to “Both roots > m”. Compared with “Both roots in (m, n)”, which condition is missing?

The graph of the quadratic function $f (x) = a x^{2} + b x + c$ ( $a \neq = 0$ ) is a parabola with axis of symmetry $x = - \frac{b}{2 a}$ . When $a > 0$ it opens upward, decreases to the left of the axis and increases to the right, and its minimum is $\frac{4 a c - b ^{2}}{4 a}$ . Asking “where are the roots of the equation?” means asking where the parabola meets the $x$ -axis. Taking $a = 1$ :

both roots in (m, n) ⟺ Δ \geq 0, m < - \frac{b}{2} < n, f (m) > 0, f (n) > 0 .

m lies between the roots ⟺ f (m) < 0 .

Predict first

$f (x) = x^{2} + b x + c$ . Do $f (0) > 0$ and $f (1) > 0$ alone guarantee that both roots of $f (x) = 0$ lie in $(0, 1)$ ?

yesno

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

Why: check the conditions one by one and see which one is guarding the gate

Example: both roots of x² − 2ax + a + 2 = 0 lie in (0, 3). Find the range of a.

Step 1 / 5

The discriminant
For $a = 1$ , $f (x) = x^{2} - 2 x + 3$ and $Δ < 0$ : the whole parabola is above the $x$ -axis. For roots we need $Δ = 4 a^{2} - 4 (a + 2) \geq 0$ , that is, $a \leq - 1$ or $a \geq 2$ . (Here we want two different roots, so we take $>$ .)

$Δ = 4 (a^{2} - a - 2) > 0 ⟺ a < - 1 or a > 2$
The axis
The right end
The left end
Intersect

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: if the axis lies in $(0, 3)$ , $f (0) > 0$ and $f (3) > 0$ , then both roots lie in $(0, 3)$

$x^{2} - 3 x + 1$ : $Δ = 5 > 0$ , the axis is $\frac{3}{2}$ , and $f (0) = f (3) = 1 > 0$ . Both roots $\frac{3 \pm 5}{2}$ lie in the interval.

Practice

Exercise 1Basic

$f (x) = x^{2} + b x + c$ satisfies $f (2 + t) = f (2 - t)$ for every real $t$ , and $f (1) = - 3$ . Find $f (4)$ .

Exercise 2Basic

Both real roots (possibly equal) of $x^{2} - m x + 4 = 0$ are greater than $1$ . Find the smallest possible $m$ .

Exercise 3Advanced

One root of $2 x^{2} - a x + 1 = 0$ lies in $(0, 1)$ and the other in $(1, 2)$ . Find the integer $a$ .

Exercise 4Challenge

Both real roots (possibly equal) of $x^{2} + (a - 3) x + a = 0$ lie in $(0, 2)$ . Find the largest possible $a$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Writing only $f(1)>0$ for “both roots greater than 1”

Both roots of

x^{2} + 4 x + 3 = 0

are greater than

1

, because

f (1) = 8 > 0

What went wrong? Think first, then open

Diagnosis $f (1) > 0$ only says that $1$ does not lie between the roots: both roots could be to the right of $1$ , or both to the left. Here the roots are $- 1$ and $- 3$ , both on the left.

You also need the axis: both roots greater than

1

⟺

Δ \geq 0

- \frac{b}{2} > 1

f (1) > 0

. Here

- \frac{b}{2} = - 2 < 1

, so it fails.

“The sign at an end point only tells “same side” or “opposite sides”; which side comes from the axis”

Exhibit

Checking the axis and the end points, forgetting the discriminant

x^{2} - 3 x + 3

: the axis

x = \frac{3}{2}

lies in

(0, 3)

f (0) = 3 > 0

and

f (3) = 3 > 0

, so both roots lie in

(0, 3)

What went wrong? Think first, then open

Diagnosis $Δ = 9 - 12 < 0$ : there are no real roots at all. The whole parabola floats above the $x$ -axis; the end-point and axis conditions hold, but there are no roots to place.

“Both roots in

(m, n)

” needs all four conditions:

Δ \geq 0

m < - \frac{b}{2} < n

f (m) > 0

f (n) > 0

“First ask whether there are roots, then ask where they are”