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8.05 The quadratic formula and the discriminant

Lesson

Concept summary

So far the methods we have looked at are useful for quadratic equations that are of a particular form, but we cannot use these methods to solve all quadratic equations. The quadratic formula is a method we can use to solve any quadratic equation written in the form ax^2+bx+c=0 and also to quickly determine the number of real solutions it has.

The quadratic formula is: x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

The expression under the radical is known as the discriminant, and we can use the sign of this value to determine the number of real solutions.

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.

Example:

5x^2-3x+9=0

Worked examples

Example 1

Use the discriminant to determine the number and nature of the solutions of the following quadratic equations:

a

2x^2-8x+3=0

Approach

The discriminant is equal to b^2-4ac, and for this equation we have a=2, b=-8, c=3.

Solution

The discriminant is (-8)^2-4(2)(3)=40, which is positive, so the equation has two real solutions.

b

-5x^2+6x-2=0

Approach

The discriminant is equal to b^2-4ac, and for this equation we have a=-5, b=6, c=-2.

Solution

The discriminant has a value of (6)^2-4(-5)(-2)=-4, which is negative, so the equation has no real solutions.

Reflection

As the quadratic formula involves taking the square root of the discriminant, if the discriminant is negative then the formula will involve taking the square root of a negative number, which will in turn result in no real solutions.

We can also see that if 4ac>b^2 the discriminant will be negative, and the corresponding equation will have no real solutions.

c

x^2-6x+9=0

Approach

The discriminant is equal to b^2-4ac, and for this equation we have a=1, b=-6, c=9.

Solution

The discriminant has a value of (-6)^2-4(1)(9)=0, so the equation has one real solution.

Reflection

If the discriminant is zero, the quadratic formula simplifies to be x=-\dfrac{b}{2a}, which is equal to the x-value of the vertex. This means the vertex lies on the x-axis and the x-coordinate of the vertex is the only solution to the quadratic equation.

Example 2

A ball is launched from a height of 80\text{ ft} with an initial velocity of 107\text{ ft} per second. Its height, h feet, after x seconds is given by h=-16x^2+107x+80Determine the number of seconds it will take the ball to reach the ground. Explain your reasoning.

Approach

On the ground the ball will have a height of 0\text{ ft}. We want to solve the equation 0=-16x^2+107x+80, and since the numbers are large the quadratic formula is an appropriate method to solve. The discriminant is equal to (107)^2-4\left(-16\right)(80)=16\,569, so we know there must be two real solutions.

Solution

Substituting into the quadratic equation we get

\dfrac{-(107) \pm \sqrt{\left(107\right)^2-4\times\left(-16\right)\times 80}}{2\times\left(-16\right)}=\dfrac{-107 \pm 3\sqrt{1841}}{-32}

The two solutions rounded to two decimal places are therefore x=-0.68,\, x=7.37. We can exclude the negative solution as it is outside the domain, which is x\geq 0.

The ball will reach the ground after 7.37 seconds.

Reflection

In most real life applications, where x is a unit of measurement, such as seconds or feet, the domain is restricted to non-negative numbers. For these types of problems we need to exclude any negative solutions.

Outcomes

MA.912.AR.1.2

Rearrange equations or formulas to isolate a quantity of interest.

MA.912.AR.3.1

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

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.1.4

Apply previous understanding of operations with rational numbers to add, subtract, multiply and divide numerical radicals.

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