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Further Quadratics
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2.0]
Discriminants of Quadratics
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2.1]
Roots of Quadratics

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2.2]
Quadratic Inequalities

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2.ECQ]
End of Chapter Questions

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2.0 Discriminants of Quadratics
Quadratics are a type of polynomial that is expressed in the form
ax2+bx+cax^2+bx+c
. They are often used to model real-world situations, such as the trajectory of a ball or the shape of a bridge. In the traditional sense, when we're told to solve a quadratic, we find the
xx
-coordinates where the curve formed crosses
y=0y=0
. However, they are actually called the roots of a quadratic. We use the below formula to determine if there are 2, 1, or 0 roots.
Note 2.0.0 - The discriminant of a quadratic
The discriminant of a quadratic function:
b2−4acb^2-4ac


1) If
b2−4ac>0b^2-4ac>0
, the roots of the equation are
unequal and real


2) If
b2−4ac=0b^2-4ac=0
, the roots of the equation are
equal and real


3) If
b2−4ac<0b^2-4ac<0
, the roots of the equation are
complex


00
b2−4ac>0b^2-4ac>0
00
b2−4ac=0b^2-4ac=0
00
b2−4ac<0b^2-4ac<0
If we analyze why some quadratic functions give us math errors on calculators (like stated above), we can come the realization that if we want to square-root a negative number, we get said math error. It holds true when we try to square-root a negative discriminant, which gives us a math error (and hence the graph at
y=0y=0
can't exist because there's simply no answer to it).
Some examples below will walk you through how to determine the amount of roots, and hence solve simple quadratics (when possible).
Example 2.0.0 - Finding the discriminant and solving equations
1) Given that
f(x)=2x2+8x+3f(x)=2x^2+8x+3
, find the value of the discriminant of
f(x)f(x)


2) Suppose
y=2x2−12x+ky=2x^2-12x+k
, where
kk
is a constant. For what value of
kk
will the equation
y=0y=0
have exactly one solution?

3) The quadratic equation
kx2+(3k−1)x−4=0kx^2+(3k-1)x-4=0
has no real roots. Find the set of possible values of
kk
1) Simply enter the values of the quadratic into their respective positions in the discriminant, giving you:
82−4×2×3=408^2-4\times2\times3=40
.

2) We know some of the values in the discriminant, and we know we need to find it when
y=0y=0
, therefore we again just enter the values and do some re-arranging:
(−12)2−4(2)(k)=0(-12)^2-4(2)(k)=0
then
8k=1448k=144
, thus we get
k=18k=18
.

3) Again, a case of substitution and re-arranging some variables. We first enter all that we know into the discriminant:
(3k−1)2−4×k×−4(3k-1)^2-4\times k\times-4
. We can then collect like terms and form a quadratic in itself to solve for
kk
:
9k2+10k+19k^2+10k+1
. As we want to get the values that make no real roots, we set the quadratic to be less than
00
:
9k2+10k+1<09k^2+10k+1<0
. We can then solve this quadratic to get the set of possible values of
kk
:
−1-1
and
−19-\frac{1}{9}
.
Now that you've had a chance to look at some simple examples, below are some further examples to test your understanding!
Problem 2.0.0 - Solving discriminant quadratics
1) Determine the number of real roots of the equation
f(x)=(x−3)(2x2+3x+5)f(x)=(x-3)(2x^2+3x+5)


2) Find the range of values of
pp
such that the equation
px2−2x+3=0,p≠0,px^2-2x+3=0, p \neq 0,
has no real roots

3) Calculate the discriminant of the equation
2x2+4x+5=02x^2+4x+5=0


4) Given that the equation
2x2−12x+k=02x^2-12x+k=0
has repeated roots, find the value of the constant
kk


5) The equation
2x2+(3p−2)x+p=02x^2+(3p-2)x+p=0
has equal roots. Determine the possible values of
pp


6) The straight line with the equation
y=7x−10py=7x-10p
does not cross or touch the curve with the equation
y=4px2−x+6y=4px^2-x+6
, where
pp
is a constant. Find the set of possible values of
pp


7) Show that the graph of the function
y=2x2−5x+6y=2x^2-5x+6
does not cross the
xx
-axis. You must use algebra to support your explanation

8) The equation
20x2=4kx−13kx2+220x^2=4kx-13kx^2+2
, where
kk
is a constant, has no real roots. Show that
kk
satisfies the inequality
2k2+ak+b<02k^2+ak+b<0
where
aa
and
bb
are constants to be found

9) Find the values of
kk
for which the equation
x2+kx+16=0x^2+kx+16=0
has no real roots

10) Find all the values of
kk
for which the equation
x2+2kx+9k=−4xx^2+2kx+9k=-4x
has two distinct real roots
2.1 Roots of Quadratics
When finding solutions to quadratic functions, we normally get two roots. Putting these roots into mathematical context, we get the following:
Note 2.1.0 - Root manipulation
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}


So then:

1) The sum of two roots is defined as:
α+β=−b+b2−4ac2a+−b−b2−4ac2a\alpha + \beta = \frac{-b+\sqrt{b^2-4ac}}{2a} + \frac{-b-\sqrt{b^2-4ac}}{2a}
, which simplifies to
−ba-\frac{b}{a}


2) The product of two roots is defined as:
αβ=−b+b2−4ac2a×−b−b2−4ac2a\alpha \beta = \frac{-b+\sqrt{b^2-4ac}}{2a} \times \frac{-b-\sqrt{b^2-4ac}}{2a}
, which simplifies to
ca\frac{c}{a}


We can use these simple rules to find root sums and products with more complex alpha-beta root expressions. Below are some examples to solidify your understanding of root manipulation.
Example 2.1.0 - Manipulating basic root expressions
1) The quadratic equation
2x2−7x−5=02x^2-7x-5=0
haas roots
α\alpha
and
β\beta
. Find:
α+β\alpha + \beta
,
αβ\alpha \beta
,
α2+β2\alpha^2 + \beta^2
, and
1α+1β\frac{1}{\alpha} + \frac{1}{\beta}


1) We first find the sum and products of the roots:
α+β=72=3.5\alpha + \beta = \frac{7}{2}=3.5
and
αβ=−52=−2.5\alpha \beta = \frac{-5}{2}=-2.5
. We can then do algebraic manipulation to find
α2+β2\alpha^2 + \beta^2
: if we expand them out, we get
(α+β)2=α2+2αβ+β2\left(\alpha + \beta \right)^2=\alpha^2+2\alpha \beta + \beta^2
. Re-arranging gives us:
α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2=\left(\alpha + \beta \right)^2-2\alpha \beta
. After that, we simply plug in the sums and products, treating them like variables:
α2+β2=(3.5)2−2×−2.5=17.25\alpha^2 + \beta^2=\left(3.5 \right)^2-2 \times -2.5=17.25
. It's the same deal with fractions:
1α+1β=β+ααβ=α+βαβ\frac{1}{\alpha} + \frac{1}{\beta}=\frac{\beta + \alpha}{\alpha \beta}=\frac{\alpha + \beta}{\alpha \beta}
. And substituting values in, we get
1α+1β=3.5−2.5=−1.4\frac{1}{\alpha} + \frac{1}{\beta}=\frac{3.5}{-2.5}=-1.4


2.2 Quadratic Inequalities
Solving inequalities with quadratics can be helpful as they can give us a large (or small) range of values that can be solutions of an equation. Answers can either be in the form
z<x<yz<x<y
or in the form
x<zx<z
and
x>yx>y
. You must be careful with wording your answer in the second form, because if you denote it as
x<z and x>yx<z \ and \ x>y
, you are saying that a single solution can be in BOTH region
zz
and
yy
, which would be a superposition of both, which in classical math is impossible!
This page was created by 9662e103-129a, and was last updated ~ June 2024