Question

Anderson uses the discriminant to correctly find the number of real solutions of the quadratic equation x2 + 4x + 8 = 0. Which explanation could Anderson provide?
The equation has no real number solutions because the discriminant is 0.
The equation has one real number solution because the discriminant is 0.
The equation has no real number solutions because the discriminant is less than 0.
The equation has two real number solutions because the discriminant is greater than 0.

Answer 1

The discriminant of ax^2+bx+c is b^2-4ac.  

If the discriminant, d, is:

d<0, there are no real solutions (there are two imaginary solutions however)

d=0, there is one real solution

d>0, there are two real solutions

In this case d=16-32=-16 so there are two imaginary solutions.  Or from your answer choices, it can be said that there are no real number solutions because the discriminant is less than zero.

Answer 2

You can use the formula for roots of a quadratic equation to find the solutions to the given quadratic equations.

The explanation that Anderson could provide is

Option C: The equation has no real number solutions because the discriminant is less than 0.

How to find the solution to a standard quadratic equation?

Suppose the given quadratic equation is $ax^2 + bx + c = 0$

Then its solutions are given as

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Using the above method to find the solutions to the given quadratic equation:

The given equation is $x^2 + 4x + 8 = 0$

Comparing it with the standard form $ax^2 + bx + c = 0$

we get: $a = 1, b = 4, c = 8$

Putting it in the aforesaid formula, we get

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \dfrac{-4 \pm \sqrt{4^2 = 4 \times 1 \times 8}}{2 \times 1} = \dfrac{-4 \pm \sqrt{-32}}{2}}{}$

Since the second term in the denominator contains negative quantity under square root, thus, both the solutions will come out to be complex numbers.

The quantity $b^2 - 4ac$ is called discriminant.

The solution contains the term $\sqrt{b^2 - 4ac}$ which will be real only if discriminant is 0 or positive. It came out to be < 0, thus, the solutions to the given quadratic equation became non real.

Thus,

The explanation that Anderson could provide is

Option C: The equation has no real number solutions because the discriminant is less than 0.

Learn more about discriminant of  quadratic equations here:

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