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# Discriminant

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Suppose you want to know whether the function $y=x^2-5x+12$ has real roots. What part of the quadratic formula would you need to test in order to determine if the roots of the function are real or complex?

### Guidance

Consider the potential roots of a quadratic function. There are three options:

1. The function has 2 distinct real roots and 2 x-intercepts.

2. The function has 1 real root (of multiplicity 2) and 1 x-intercept.

3. The function has 0 real roots, 2 complex roots, and 0 x-intercepts.

The quadratic formula states that the two roots of a quadratic function $y=ax^2+bx+c$ are:

$r_1= \frac{-b+ \sqrt{b^2-4ac}}{2a}$ and $r_2= \frac{-b- \sqrt{b^2-4ac}}{2a}$

Under the radical sign is the expression $b^2-4ac$ . This expression is known as the discriminant of the quadratic function and is important especially because when it is negative the roots of the function are complex. There are only three possible outcomes for the value of the discriminant. These outcomes are:

1. The value of the discriminant is positive ( $b^2-4ac>0$ ). This means the function has 2 distinct real roots and 2 x-intercepts.
2. The value of the discriminant is zero ( $b^2-4ac=0$ ). This means the function has 1 real root (of multiplicity 2) and 1 x-intercept.
3. The value of the discriminant is negative ( $b^2-4ac<0$ ). This means the function has 0 real roots, 2 complex roots, and 0 x-intercepts.

This discriminant is helpful when you are only looking to describe the graph of a quadratic function or its roots, but don't need to know its exact roots. If you need to know its exact roots, you will still have to use the complete quadratic formula.

#### Example A

If the discriminant of a quadratic function has the value shown below, determine if the function will have two distinct real roots, 1 real root of multiplicity 2, or two distinct complex roots.

a) 7

b) –3

c) $\frac{1}{2}$

d) 0

Solution:

a) $b^2-4ac>0$ so the quadratic function will have two distinct real roots.

b) $b^2-4ac<0$ so the quadratic function will have two distinct complex roots.

c) $b^2-4ac>0$ so the quadratic function will have two distinct real roots.

d) $b^2-4ac=0$ so the quadratic function will have 1 real root of multiplicity 2.

#### Example B

Given the function $y = ax^2+ bx + c$ , how many times will the graph intersect the $x$ -axis if the discriminant of its corresponding quadratic equation has the value:

a) 5

b) –6

c) 0

d) 0.2

Solution:

a) $b^2-4ac>0$ so the parabola will cross the $x$ -axis twice.

b) $b^2-4ac<0$ so the parabola will not cross the $x$ -axis.

c) $b^2-4ac=0$ so the parabola will cross the $x$ -axis once.

d) $b^2-4ac>0$ so the parabola will cross the $x$ -axis twice.

#### Example C

For each of the following quadratic equations, determine the value of the discriminant and use that value to describe the nature of the roots.

a) $4x^2-4x+1=0$

b) $x^2+3x+9=0$

c) $2x^2+3x-4=0$

d) $9x^2+12x+4=0$

Solution: Let $D$ represent the discriminant.

a) The quadratic equation will have 1 real solution of multiplicity 2.

$& D=b^2-4ac \\& D=({\color{red}{-4}})^2-4({\color{red}{4}})({\color{red}{1}}) \\& D=16-16 \\&\boxed{D=0} \\& b^2-4ac=0$

b) The quadratic equation will have two distinct complex solutions and no real solutions.

$& D=b^2-4ac \\& D=({\color{red}{3}})^2-4({\color{red}{1}})({\color{red}{9}}) \\& D=9-36 \\&\boxed{D=-27} \\& b^2-4ac<0$

c) The quadratic equation will have two distinct real solutions.

$& D=b^2-4ac \\& D=({\color{red}{3}})^2-4({\color{red}{2}})({\color{red}{-4}}) \\& D=9+32 \\&\boxed{D=41} \\& b^2-4ac>0$

d) The quadratic equation will have 1 real root of multiplicity 2.

$& D=b^2-4ac \\& D=({\color{red}{12}})^2-4({\color{red}{9}})({\color{red}{4}}) \\& D=144-144 \\&\boxed{D=0} \\& b^2-4ac=0$

#### Concept Problem Revisited

In order to determine if the roots of $y=x^2-5x+12$ are real or complex, you must find the discriminant, $b^2-4ac$ . For this function:

$b^2-4ac&=(-5)^2-4(1)(12)\\&=25-48\\&=-23$

Because the discriminant is negative, this function has two distinct complex roots.

### Vocabulary

Discriminant
The discriminant is the radicand of the quadratic formula and its value is used to describe the nature of the roots of a quadratic equation.
$b^2-4ac>0 \rightarrow$ 2 distinct real roots
$b^2-4ac=0 \rightarrow$ 1 real root of multiplicity 2
$b^2-4ac<0 \rightarrow$ 2 distinct complex roots and 0 real roots
Multiplicity
The multiplicity of a root or solution is the number of times that the same solution is obtained through factoring or the quadratic formula. When the discriminant is equal to zero, both solutions produced by the quadratic formula will be the same. There is only one solution, but it is said to have "multiplicity 2".

### Guided Practice

1. Given the following quadratic equation, find the value of ‘ $m$ ’ such that the equation will have 1 real solution of multiplicity 2.

$mx^2+(m+8)x+9=0$

2. Given the following quadratic equation, determine the nature of the solutions:

$4(y^2-5y+5)=-5$

3. Given the following quadratic equation, find the value of ‘ $m$ ’ such that the equation will have two distinct complex solutions.

$(m+1)e^2-2e-3=0$

1. Begin by determining the value of the discriminant.

$b^2-4ac=({\color{red}{m+8}})^2-4({\color{red}{m}})({\color{red}{9}})$

Expand and Simplify

$& b^2-4ac=({\color{red}{m+8}})({\color{red}{m+8}})-4({\color{red}{m}})({\color{red}{9}}) \\& b^2-4ac=m^2+8m+8m+64-36m \\& b^2-4ac=m^2-20m+64$

If the equation has 1 real solution of multiplicity 2, the value of the discriminant must equal zero.

$& b^2-4ac=0 \\& \therefore m^2-20m+64=0$

Factor the quadratic equation and solve for the variable ‘ $m$ ’.

$&(m-16)(m-4)=0 \\& m-16=0 \ or \ m-4=0 \\& m={\color{red}{16}} \ or \ m={\color{red}{4}}$

The values of ‘ $m$ ’ that would produce 1 solution of multiplicity 2 for the quadratic equation are $m=16 \ or \ m=4$ .

2. Write the quadratic equation in general form.

$4(y^2-5y+5)=-5$

Apply the distributive property.

$4y^2-20y+20=-5$

Set the equation equal to zero.

$&4y^2-20y+20{\color{red}{+5}}=-5{\color{red}{+5}} \\&4y^2-20y+25=0$

Determine the value of the discriminant for this quadratic equation.

$& D=b^2-4ac \\& D=({\color{red}{-20}})^2-4({\color{red}{4}})({\color{red}{25}}) \\& D=400-400 \\&\boxed{D=0} \\& b^2-4ac=0$

The quadratic equation will have 1 real solution of multiplicity 2.

3. Begin by determining the value of the discriminant.

$b^2-4ac=({\color{red}{-2}})^2-4({\color{red}{m+1}})({\color{red}{-3}})$

Expand and Simplify

$b^2-4ac=12m+16$

If the equation has two distinct complex solutions the value of the discriminant must be less than zero.

$& b^2-4ac<0 \\& \therefore 12m+16<0$

Solve the inequality.

$&12m+16{\color{red}{-16}}<0{\color{red}{-16}} \\&12m<-16 \\&\frac{12m}{{\color{red}{12}}}< \frac{-16}{{\color{red}{12}}} \\&\frac{\cancel{12}m}{\cancel{12}}<- \frac{16}{12} \\&\boxed{m<- \frac{4}{3}}$

The value of ‘ $m$ ’ that would produce two distinct complex solutions for the quadratic equation is $m<- \frac{4}{3}$ .

### Practice

If the discriminant of a quadratic equation has the value shown below, describe the nature of the solutions.

1. –14
2. 11
3. 0
4. –0.25
5. 124

State the nature of the solutions for each of the following quadratic equations.

1. $2x^2+7x-1=0$
2. $3x^2+2x=-7$
3. $-9x^2-7=4x$
4. $x^2-8x+16=0$
5. $4+2x^2=11x$

Determine the value(s) of ‘ $m$ ’ that will produce the indicated solution for each of the following:

1. $y^2+(m+2)y+2m=0$ ; 1 real solution of multiplicity 2
2. $g^2+(m-1)g+1=0$ ; 2 real solutions
3. $3mx^2-3x+1=0$ ; 2 complex solutions
4. $x^2+4mx+1=0$ ; 1 real solution of multiplicity 2
5. $p^2+mp+16=0$ ; 2 real solutions