Topics
2. Quadratic Functions & Equations
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United States of AmericaFL
Grade 912

2.04 The quadratic formula & discriminant

Lesson
Practice
Lesson

Concept summary

We have seen that not all quadratic equations have real solutions. We cannot determine the complex root (of a quadratic) by graphing, but we can using the quadratic formula.

\displaystyle x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
\bm{b^2-4ac}
the discriminant

We can use the discriminant to determine the nature of the solutions, and also the value of the solutions whether they are real or complex.

b^2-4ac>0

The equation has two real solutions

Example:

x^2-4x-8=0

b^2-4ac=0

The equation has one real solution

Example:

9x^2+6x+1=0

b^2-4ac<0

The equation has no real solutions and two complex solutions

Example:

5x^2-3x+9=0

b^2-4ac=k^2

Where a, b, c, and k are rational numbers. The equation has rational solutions.

Example:

x^2+5x+6=0

Worked examples

Example 1

Calculate the value of the discriminant for the following equations and use it to determine the number and nature of the solutions:

a

5x^2+2x+2=0

Approach

The discriminant is equal to b^2-4ac, and for this equation we have a=5, b=2, c=2.

Solution

The discriminant is (2)^2-4(5)(2)=-36, so no real solutions and two complex solutions.

Reflection

If the discriminant is negative, the quadratic formula needs to find the square root of a negative number, which results in non-real solutions. We can see that if 4ac>b^2 the discriminant will always be negative.

b

16x^2-24x+9=0

Approach

The discriminant is equal to b^2-4ac, and for this equation we have a=16, b=-24, c=9.

Solution

The discriminant is (-24)^2-4(16)(9)=0. Notice that all the coefficients are rational and 0^2=0, so one real rational solution.

Reflection

If the discriminant is zero, there is only one real solution. It is not possible to have one real solution and one complex solution for a quadratic equation with real coefficients.

Example 2

Solve the following equations, stating your solutions in the form a \pm b i:

a

x^{2} - 6 x + 19 = 0

Approach

A quadratic equation in standard form ax^2+bx+c=0 has the solutions x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}, and for this equation we have a=1, b=-6, c=19.

Solution

\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle =\displaystyle \frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\left( 1\right)\left( 19\right)}}{2\left( 1\right)}Substitute a=1, b=-6, c=19
\displaystyle =\displaystyle \frac{6\pm \sqrt{36-76}}{2}Evaluate the square and products
\displaystyle =\displaystyle \frac{6\pm \sqrt{-40}}{2}Evaluate the difference in the radicand
\displaystyle =\displaystyle \frac{6\pm \sqrt{40i^2}}{2}Rewrite the radicand as a complex number
\displaystyle =\displaystyle \frac{6\pm 2\sqrt{10}i}{2}Simplify the radical
\displaystyle =\displaystyle \frac{6}{2}\pm \frac{2\sqrt{10}i}{2}Rewrite as two fractions
\displaystyle =\displaystyle 3\pm \sqrt{10}iSimplify the quotients

Reflection

We can see that the two complex solutions, 3+\sqrt{10}i and 3-\sqrt{10}i, are complex congugates. That is, they are of the form a+bi and a-bi.

b

4x^2+9=0

Approach

A quadratic equation in standard form ax^2+bx+c=0 has the solutions x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}, and for this equation we have a=4, b=0, c=9.

Solution

\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle =\displaystyle \frac{-0\pm \sqrt{0^2-4\left( 4\right)\left( 9\right)}}{2\left( 4\right)}Substitute a=4, b=0, c=9
\displaystyle =\displaystyle \frac{\pm \sqrt{0-144}}{8}Evaluate the square and products
\displaystyle =\displaystyle \frac{\pm \sqrt{-144}}{8}Evaluate the difference in the radicand
\displaystyle =\displaystyle \frac{\pm \sqrt{144i^2}}{8}Rewrite the radicand as a complex number
\displaystyle =\displaystyle \frac{\pm 12i}{8}Simplify the radical
\displaystyle =\displaystyle \pm \frac{3}{2}iSimplify the quotient

Reflection

If the discriminant is zero, there is only one real solution. It is not possible to have one real solution and one complex solution for a quadratic equation with real coefficients.

\pm \dfrac{3}{2}i is in the form a\pm bi with a=0.

Example 3

Consider the quadratic function p\left(x\right)=x^2-6x+16.

a

Find the roots of the equation p \left( x \right) = 0.

Approach

The roots of the equation p \left( x \right) = 0 are the same as the solutions to the equation, so we can use the quadratic formula to find the roots.

Solution

\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle =\displaystyle \frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\left( 1\right)\left( 16\right)}}{2\left( 1\right)}Substitute a=1, b=-6, c=16
\displaystyle =\displaystyle \frac{6\pm \sqrt{-28}}{2}Evaluate the radicand
\displaystyle =\displaystyle \frac{6\pm 2\sqrt{-7}}{2}Simplify the radicand
\displaystyle =\displaystyle 3\pm\sqrt{-7}Evaluate the quotient
\displaystyle =\displaystyle 3\pm\sqrt{7}iRewrite the radicand as a complex number

Reflection

In this example we simplified the two solutions, leaving the conversion of the negative radicand until the last step. It does not matter which order you perform these operations, the result will be the same.

b

Determine the nature of these roots.

Solution

The roots are complex roots as they are in the form a\pm bi.

Reflection

Without calculating the value of the roots, we can determine they are complex roots as we know the discriminant is less than zero.

Outcomes

MA.912.AR.3.2

Given a mathematical or real-world context, write and solve one-variable quadratic equations over the real and complex number systems.

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

MA.912.NSO.2.1

Extend previous understanding of the real number system to include the complex number system. Add, subtract, multiply and divide complex numbers.

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