2.0 Discriminants of Quadratics

Quadratics are a type of polynomial that is expressed in the form

$ax^2+bx+c$

. 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 $x$

-coordinates where the curve formed crosses $y=0$

. 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: $b^2-4ac$

1) If

$b^2-4ac>0$

, the roots of the equation are unequal and real

Two roots (x-intercepts)

2) If

$b^2-4ac=0$

, the roots of the equation are equal and real

One root (x-intercepts)

3) If

$b^2-4ac<0$

, the roots of the equation are complex

No real roots (x-intercepts)

$b^2-4ac>0$

$b^2-4ac=0$

$b^2-4ac<0$

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

$y=0$

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 $f(x)=2x^2+8x+3$

, find the value of the discriminant of $f(x)$

2) Suppose

$y=2x^2-12x+k$

, where $k$

is a constant. For what value of $k$

will the equation $y=0$

have exactly one solution?3) The quadratic equation

$kx^2+(3k-1)x-4=0$

has no real roots. Find the set of possible values of $k$

1) Simply enter the values of the quadratic into their respective positions in the discriminant, giving you: $8^2-4\times2\times3=40$

.2) We know some of the values in the discriminant, and we know we need to find it when

$y=0$

, therefore we again just enter the values and do some re-arranging: $(-12)^2-4(2)(k)=0$

then $8k=144$

, thus we get $k=18$

.3) Again, a case of substitution and re-arranging some variables. We first enter all that we know into the discriminant:

$(3k-1)^2-4\times k\times-4$

. We can then collect like terms and form a quadratic in itself to solve for $k$

: $9k^2+10k+1$

. As we want to get the values that make no real roots, we set the quadratic to be less than $0$

: $9k^2+10k+1<0$

. We can then solve this quadratic to get the set of possible values of $k$

: $-1$

and $-\frac{1}{9}$

.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 $f(x)=(x-3)(2x^2+3x+5)$

2) Find the range of values of

$p$

such that the equation $px^2-2x+3=0, p \neq 0,$

has no real roots3) Calculate the discriminant of the equation

$2x^2+4x+5=0$

4) Given that the equation

$2x^2-12x+k=0$

has repeated roots, find the value of the constant $k$

5) The equation

$2x^2+(3p-2)x+p=0$

has equal roots. Determine the possible values of $p$

6) The straight line with the equation

$y=7x-10p$

does not cross or touch the curve with the equation $y=4px^2-x+6$

, where $p$

is a constant. Find the set of possible values of $p$

7) Show that the graph of the function

$y=2x^2-5x+6$

does not cross the $x$

-axis. You must use algebra to support your explanation8) The equation

$20x^2=4kx-13kx^2+2$

, where $k$

is a constant, has no real roots. Show that $k$

satisfies the inequality $2k^2+ak+b<0$

where $a$

and $b$

are constants to be found9) Find the values of

$k$

for which the equation $x^2+kx+16=0$

has no real roots10) Find all the values of

$k$

for which the equation $x^2+2kx+9k=-4x$

has two distinct real roots2.1 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

$\alpha = \frac{-b+\sqrt{b^2-4ac}}{2a}$

and $\beta = \frac{-b-\sqrt{b^2-4ac}}{2a}$

So then:

1) The sum of two roots is defined as:

$\alpha + \beta = \frac{-b+\sqrt{b^2-4ac}}{2a} + \frac{-b-\sqrt{b^2-4ac}}{2a}$

, which simplifies to $-\frac{b}{a}$

2) The product of two roots is defined as:

$\alpha \beta = \frac{-b+\sqrt{b^2-4ac}}{2a} \times \frac{-b-\sqrt{b^2-4ac}}{2a}$

, which simplifies to $\frac{c}{a}$

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 $2x^2-7x-5=0$

haas roots $\alpha$

and $\beta$

. Find: $\alpha + \beta$

, $\alpha \beta$

, $\alpha^2 + \beta^2$

, and $\frac{1}{\alpha} + \frac{1}{\beta}$

1) We first find the sum and products of the roots:

$\alpha + \beta = \frac{7}{2}=3.5$

and $\alpha \beta = \frac{-5}{2}=-2.5$

. We can then do algebraic manipulation to find $\alpha^2 + \beta^2$

: if we expand them out, we get $\left(\alpha + \beta \right)^2=\alpha^2+2\alpha \beta + \beta^2$

. Re-arranging gives us: $\alpha^2 + \beta^2=\left(\alpha + \beta \right)^2-2\alpha \beta$

. After that, we simply plug in the sums and products, treating them like variables: $\alpha^2 + \beta^2=\left(3.5 \right)^2-2 \times -2.5=17.25$

. It's the same deal with fractions: $\frac{1}{\alpha} + \frac{1}{\beta}=\frac{\beta + \alpha}{\alpha \beta}=\frac{\alpha + \beta}{\alpha \beta}$

. And substituting values in, we get $\frac{1}{\alpha} + \frac{1}{\beta}=\frac{3.5}{-2.5}=-1.4$

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

$z<x<y$

or in the form $x<z$

and $x>y$

. You must be careful with wording your answer in the second form, because if you denote it as $x<z \ and \ x>y$

, you are saying that a single solution can be in BOTH region $z$

and $y$

, 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