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Δβα\Delta _{\beta \alpha}
01
Linear equations
Δβα\Delta _{\beta \alpha}
02
Quadratic equations
Δβα\Delta _{\beta \alpha}
03
Functions
Δβα\Delta _{\beta \alpha}
04
Linear algebra
Δβα\Delta _{\beta \alpha}
02
Quadratic equations
Starter
List out everything you know about quadratic equations. This can be things like the formula, or if you already know the concept of a discriminant, that too!
Introduction to the discriminant
You should already know the different forms you can express a quadratic in which can provide different types of solutions (i.e.
f(x)=ax2+bx+cf(x)=ax^2+bx+c
for roots,
f(x)=a(x−r1)(x−r2)f(x)=a(x-r_1)(x-r_2)
again for roots, and
f(x)=a(x−h)2+k)f(x)=a(x-h)^2+k)
for vertex points), so I won't bother explaining each, and instead we'll jump straight to the interesting bits: roots. Roots are the
xx
-intercepts of a polynomial function (such as a quadratic). Within a quadratic, you can have either 2, 1, or 0 roots depending on the nature of the function. We can find the number of these roots in a given quadratic through what's called the discriminant:
b2−4acb^2-4ac
. The note below describes the solutions and their cases.
Note N02.0a - Solution occurrences
b2−4ac<0b^2-4ac<0
when the quadratic has
00
real solutions (all solutions are imaginary)
b2−4ac=0b^2-4ac=0
when the quadratic has
11
real solution
b2−4ac>0b^2-4ac>0
when the quadratic has
22
distinct real solutions
Think about it, when you get a math error on your calculator when typing a quadratic into the formula
−b±b2−4ac2a\frac{-b \pm \sqrt{b^2-4ac}}{2a}
, chances are, the discriminant (the
b2−4acb^2-4ac
bit), is
<0<0
. You'll learn what happens when you square-root a negative number, in a later chapter. For now, just assume that you get no real solution (i.e. no
xx
-intercepts) when the discriminant is less than 0.
Using the
aa
coefficient and discriminant
When sketching a quadratic, we can use the discriminant to determine how many intercepts we have, but we can also use the
aa
coefficient (that's negative or positive) to determine the "directionality" that the quadratic points. If the coefficient of that first
xx
is negative, the graph points downwards i.e.
−x2+2x+1-x^2+2x+1
(go ahead and plot this for yourself). If that first variable is positive, the graph points upwards i.e.
x2+2x+1x^2+2x+1
. This should be common knowledge, hence this section is short.
Axis of symmetry
A quadratic will always have an axis of symmetry. It's just the nature of this type of function. When we look at a graph of
y=x2y=x^2
, we see that the minimum (or maximum if that
aa
coefficient is negative) is the axis of symmetry (or, more precisely, the point of symmetry). If that's the case, we can find the axis from these three forms of a quadratic:
Note N02.1a - Axis of symmetry
In the form
y=a(x−p)(x−q)y=a(x-p)(x-q)
, the axis of symmetry is:
x=p+q2x=\frac{p+q}{2}
therefore, the vertex is
(p+q2,f(p+q2))\left(\frac{p+q}{2}, f\left(\frac{p+q}{2} \right) \right)
In the form
y=a(x−h)2+ky=a(x-h)^2+k
, the axis of symmetry is:
x=hx=h
therefore, the vertex is
(h,k)\left(h,k \right)
In the form
y=ax2+bx+cy=ax^2+bx+c
, the axis of symmetry is:
x=−b2ax=-\frac{b}{2a}
therefore, the vertex is
(−b2a,f(−b2a))\left(-\frac{b}{2a}, f\left(-\frac{b}{2a} \right) \right)
It's not a smart idea to memorize these formulas, simply because they can be derived easily from the graph. We know since the graph is
symmetrical
, we can just get the average of the
xx
-values of the roots.
Positive and negative definite quadratics
The last thing you'll need to know for graphing, is whether a quadratic might be positive or negative definite. These terms essentially mean whether the quadratic ever passes the
xx
-axis in either the negative or positive
yy
Cartesian space.
Note N02.2a - Definite quadratics
Positive definite when
a>0a>0
and
Δ<0\Delta <0
for all values
x∈Rx \in \mathbb{R}
Negative definite when
a<0a<0
and
Δ<0\Delta <0
for all values
x∈Rx \in \mathbb{R}
So what?
Knowing the above information is useful when sketching these graphs. When you sketch one, you generally have to add on the roots, axis of symmetry, vertex, and
yy
-intercept to get all the marks on a question.
Coursework C02.0a - Sketching quadratic graphs
1) Sketch these quadratics, including all points mentioned above:
y=4x2−5x+2y=4x^2-5x+2
,
y=2(x−4)(x−3)y=2(x-4)(x-3)
, and
y=−3(x+2)2+1y=-3(x+2)^2+1
2) Explain why this quadratic has no real solutions:
20x2−56x+4020x^2-56x+40
3) State the values of
kk
for which this quadratic
5x2+kx−35x^2+kx-3
will have: 2 solutions, 1 solution, and 0 solutions
4) Explain why
3x2+kx−13x^2+kx-1
is never positive definite for any value of
kk
5) Find the value of
kk
such that
y=12x2+(k−2)x+k2+4y=\frac{1}{2}x^2+(k-2)x+k^2+4
is not positive definite. What relationship does the graph have with the
xx
-axis in this case?
Roots
As described earlier, the solutions of a quadratic are know as it's roots. We can construct new quadratic equations or find the sums/products of roots using these "equations":
Note N02.3a - Roots
The roots of
ax2+bx+cax^2+bx+c
are:
α=−b+b2−4ac2a\alpha=\frac{-b +\sqrt{b^2-4ac}}{2a}
and
β=−b−b2−4ac2a\beta=\frac{-b -\sqrt{b^2-4ac}}{2a}
The sum of these roots are:
α+β=−b+b2−4ac2a+−b−b2−4ac2a≡−ba\alpha + \beta=\frac{-b +\sqrt{b^2-4ac}}{2a}+\frac{-b -\sqrt{b^2-4ac}}{2a} \equiv -\frac{b}{a}
The product of these roots are:
αβ=(−b+b2−4ac2a)(−b−b2−4ac2a)≡ca\alpha \beta=\left( \frac{-b +\sqrt{b^2-4ac}}{2a} \right) \left( \frac{-b -\sqrt{b^2-4ac}}{2a} \right) \equiv \frac{c}{a}
A key form that would be good to memorize is:
(α+β)2≡α2+β2+2αβ\left(\alpha + \beta \right)^2 \equiv \alpha ^2 + \beta ^2 + 2\alpha \beta
, and you can just re-arrange when you need to find
α2+β2\alpha ^2 + \beta ^2
Simple algebraic manipulation can be used to solve questions related to roots. The following example shows how a typical question is solved:
Example E02.0a - Solving roots
The quadratic equation
x2−2kx+(k−1)=0x^2-2kx+(k-1)=0
has roots
α\alpha
and
β\beta
such that
α2+β2=4\alpha ^2 + \beta^2=4
. Without solving the equation, find the possible values of the real number
kk
.
First, find
α+β=2k\alpha + \beta=2k
and
αβ=k−1\alpha \beta=k-1
Expand the roots:
(α+β)2=α2+αβ+αβ+β2\left(\alpha + \beta \right)^2=\alpha ^2+\alpha \beta + \alpha \beta + \beta ^2
Simplify and substitute values:
(α+β)2=(α2+β2)+2αβ\left(\alpha + \beta \right)^2=\left(\alpha ^2 + \beta ^2 \right) + 2\alpha \beta
We know that
α2+β2=4\alpha ^2 + \beta^2=4
, so:
4k2=4+2(k−1)4k^2=4+2(k-1)
Simplify and re-arrange:
4k2−2k−2=04k^2-2k-2=0
then
2k2−k−1=02k^2-k-1=0
then
(2k+1)(k−1)=0(2k+1)(k-1)=0
So
k=−12k=-\frac{1}{2}
or
k=1k=1