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Powerful Quadratics

This comes in two parts, with the first being less fiendish than the second. It’s great for practising both quadratics and laws of indices, and you can get a lot from making sure that you find all the solutions. For a real challenge (requiring a bit more knowledge), you could consider finding the complex solutions.

Factorisable Quadratics

This will encourage you to think about whether all quadratics can be factorised and to develop a better understanding of the effect that changing the coefficients has on the factorised form.

Which Quadratic?

In this activity you will need to work in a group to connect different representations of quadratics.

Discriminating

Age 16 to 18

Dave, from Burnham Grammar School, got us started by showing that the seventh statement is never true. Here is his solution:

If $c=0$ then $ax^2 + bx = 0$ where $a \neq 0$.
It then follows that this equation can be factorized to give $x(ax+b) = 0$.
What we then see is that the curve has a root at $x = 0$ and at $x = -b/a$, so it always has at least one real root.

He also gave an alternative way of thinking about this using the discriminant:

If $c=0$, then the discriminant is $b^2 - 4ac = b^2 - 4a(0) = b^2$.

From that we have seen the discriminant is always positive when $b \neq 0$ and so the quadratic $ax^2 + bx = 0$ always has two real distinct roots.

However, if $b = 0$ then the quadratic becomes $ax^2 = 0$, and the discriminant would be : $b^2 - 4ac = (0)^2 - 4a(0) = 0$.
As the discriminant would always be 0, the quadratic $ax^2 = 0$ will always have one repeated root.


Well done to Julian from British School Manila and Alexander who managed to complete all the questions. All the answers were correct except for subtle points in parts 4 and 11 (see the editorial comments below). Here are Alexander's answers and very clear explanations. To start with he has collected together some of the facts about the discriminant he found useful to refer to in his answers:

Before we start to answer the main statements, let us prove a few useful
facts first:

$ax^2+bx+c = 0$
$⇔ x^2+\frac{bx}{a}+\frac{c}{a} = 0$
$⇔ x^2+\frac{bx}{a} = \frac{-c}{a} (as a \neq 0)$
$⇔ (x+\frac{b}{2a})^2-\frac{b^2}{(4a^2)} = \frac{-4ac}{(4a^2)}$
$⇔ (x+\frac{b}{2a})^2= \frac{b^2-4ac}{4a ²}$
$⇔ x+\frac{b}{2a} = ±\frac{\sqrt{b^2-4ac}}{2a}$
$⇔ x = \frac{(b ±\sqrt{b^2-4ac})}{2a}$ (i)

$⇒$If $b^2-4ac > 0$,
Then it must have two square roots, and therefore the equation must have
exactly two real roots. (ii)

$⇒$If $b^2-4ac = 0$,
Then it must have one square root (0), and therefore the equation must have
exactly one (repeated) real root. (iii)

$⇒$If $b^2-4ac < 0$,
Then it must have no square roots, and therefore the equation must have
exactly zero real roots. (iv)

Let us refer to ($b^2-4ac$) as the discriminant of an equation.

Here are his answers.

(1) SOMETIMES

I give two examples:

$-x^2-3x-2 = 0$
$⇒$ The discriminant = $(-3)^2-(4 \times (-1) \times (-2))= 9-8= 1 > 0$
So through (ii), we see that this equation has two real roots.

$-x^2-2x-100$
$⇒$The discriminant $= (-2)^2-(4 \times (-1) \times (-100)) = 4-400 = -396 < 0$
So through (iv), we see that this equation has no real roots.

Using (iv), we see that if $a < 0$, the equation has no real roots if and
only if $b^2-4ac < 0$.

(2) ALWAYS

Through (iii), we see that if $b^2-4ac = 0$, then the equation has one
(repeated) real root.

(3) ALWAYS

The discriminant of the equation $ax^2+bx+c$ is $b^2-4ac$.

Through (iv), we can easily say that given that the equation has no real
roots, then $b^2-4ac < 0$.

$b^2-4ac < 0$
$⇒ b^2 < 4ac$
But as $b^2$ is always non-negative, $4ac$ must therefore be positive.

The discriminant of the equation $ax^2+bx-c$ is
$b^2-4a \times(-c) = b^2+4ac$
And as we know already that $4ac$ is positive, and $b^2$ is non-negative,
$b^2+4ac$ must be positive,
$⇒$ The discriminant of the equation $ax^2+bx-c$ is positive,
$⇒$ Through (ii), the equation $ax^2+bx-c$ must have two distinct real
roots.

(4) NEVER

If $\frac{b^2}{a} < 4c$
Then $b^2 < 4ac$
$⇒ b^2-4ac < 0$
$⇒$ Through (iv), the equation has no real roots.

In fact this statement is sometimes true; for example the quadratic $-x^2+1$ with $a=-1,b=0,c=1$ has $\frac{b^2}{a}=0<4=4c$ and yet $1$ is a real root. The reason the proof didn't work is because in the first step the inequality was multiplied through by $a$, and since in this case $a$ is a negative number it would reverse the sign of the inequality.

(5) SOMETIMES

The discriminant of an equation is $b^2-4ac$.
However, if $b = 0$,
Then $b^2-4ac= 0-4ac= -4ac$

We can easily give an example of an equation which has two real roots,
where $b = 0$:
$x^2-9 = 0$
$⇒ b^2-4ac = (0) ²-4\times (1) \times (-9)= 0+9= 9 > 0$
Through (ii), the equation must have two real roots.

Let us now find a situation where the equation has exactly one (repeated)
real root.
Through (iii), the equation has one real root if and only if the
discriminant is equal to 0.
$⇒ -4ac = 0$
$⇒ ac = 0$
$⇒$ Either one or both of a and c are equal to 0.
Obviously $a \neq 0$, otherwise the equation would not be quadratic.
$⇒ c = 0$
$⇒ b = c = 0$
Therefore, given that $b = 0$, the only equations that have one repeated real
root are of the form $ax^2 = 0$. Interestingly, this means that the solutions
to these forms of of equations are always $x = 0$.

(6) NEVER

As shown above, the only three possibilities for the number of real roots
that an equation can have are 0, 1 and 2. If an equation had three real
roots, then the discriminant would have to have three real square roots,
which is impossible.

(7) NEVER

The discriminant of an equation is $b^2-4ac$.

If $c = 0$,
Then $b^2-4ac = b^2-4a \times (0) = b^2$
Through (iv), an equation can have no real roots if and only if the
discriminant is negative.
But as $b^2 ≥ 0$ (as squares are always non-negative),
This can never happen.
Therefore if $c = 0$, then the equation always has at least one real root.

(8) ALWAYS

The discriminant of the equation $ax^2+bx+c = 0$ is $b^2-4ac$.
The discriminant of the equation $ax^2-bx+c = 0$ is $(-b)^2-4ac$
$=b ²-4ac$.

Therefore, as the two equations have the same discriminant, they must also
have the same number of real roots (through (ii), (iii) and (iv))

(9) ALWAYS

Through (ii), if an equation has two real roots, then the discriminant
$b^2-4ac > 0$.

$b^2-4ac > 0$
$⇒ b^2 > 4ac$
$⇒ ac < \frac{b^2}{4}$

Therefore this condition must always be true if an equation has two real
roots.

(10) SOMETIMES

I shall give two examples:

$x^2+3x+2 = 0$
$⇒$ The discriminant $= (3)^2-4 \times (1) \times (2)= 9-8= 1 > 0$
Therefore, through (ii), the equation has two distinct real roots.

$x^2+6x+9 = 0$
$⇒$ The discriminant $= (6)^2-4 \times (1) \times (9)= 36-36= 0$
Therefore, through (iii), the equation has exactly one real root.

The equation has two distinct real roots if and only if the discriminant,
$b^2-4ac$, is positive.

(11) ALWAYS

The discriminant of the equation $ax^2+bx+c = 0$ is $b^2-4ac$.
The discriminant of the equation $cx^2+bx+a = 0$ is $b^2-4ca$.

But $b^2-4ca = b^2-4ac$,
$⇒$ The equations have the same discriminant.
$⇒$ The equations must have the same number of real roots (through (ii),
(iii), (iv))

In fact this statement is only sometimes true because of the fact that you may not get a quadratic when you swap $a$ and $c$ around since $c$ can be $0$. So for example the equation $x^2+x=0$ has two solutions ($0$ and $-1$) but the equation $x+1=0$ which has $a$ and $c$ reversed, only has one solution ($-1$).

(12) NEVER

$-ax^2-bx-c = 0$
$⇒ ax^2+bx+c = 0$

Therefore, if
$ax^2+bx+c = 0$ has no real roots,
Then
$-ax^2-bx-c = 0$ has no real roots also.


Julian and Classmates also came up with their own discriminating question

(13) Trees with negative discriminants are able to stand.

Answer:

NEVER TRUE
They have no real roots.