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Common Core Standards: Math

High School: Number and Quantity

The Complex Number System HSN-CN.C.7

7. Solve quadratic equations with real coefficients that have complex solutions.

When students dealt with parabolas, they only saw the good parabolas. The ones with nice simple roots like -1 and 1. The roots could be found just by looking at the graph or, if they were ugly, by using the quadratic formula.

They probably didn't think too much about those "other" parabolas. The parabolas from the wrong side of the tracks. Yeah, we mean the ones that don't cross the x-axis at all. Your students didn't consider that those parabolas have roots (and feelings), too.

Well, it's time. They're old enough to know the truth.

Those quadratic equations have imaginary roots. Now that your students know that "imaginary" doesn't mean "make believe," we can make it up to these forgotten, mistreated parabolas. Flowers and chocolate just won't cut it; we'll have to work with them.

Let's take a look at one: x2 + 4x + 6 = 0. This one doesn't factor, so we'll have to use the quadratic formula:

 

For this particular parabola, our values are a = 1, b = 4 and c = 6. When we plug in those numbers, we get 

But wait. We can't have the square root of a negative number if we're working with real numbers. Good thing we're working with imaginary numbers, then, isn't it?

That means we can turn that fraction into  which simplifies to . So our parabola has 2 perfectly lovely roots. The only problem is that the graph doesn't cross the x-axis because it only contains real numbers, and the roots are imaginary, not real.

Quadratics with imaginary roots aren't always easy to identify until we get to the point in the problem where we have to take the square root. At that point, if the number under the radical is negative, that parabola's roots will be imaginary. (That number under the radical, b2 – 4ac, is called the discriminant. Probably because it discriminates between real and imaginary roots.)

Now, your students should be able to calculate the roots of these other parabolas and tell whether or not a parabola will have real or imaginary roots by looking at its discriminant.

 

Drills

  1. Does the parabola with equation x2 + 7x – 3 = 0 have real or imaginary roots?

    Correct Answer:

    Real, because the discriminant is positive

    Answer Explanation:

    To find out whether a parabola has real or imaginary roots, all we need to do is calculate the discriminant, or the part of the quadratic formula under the radical (b2 – 4ac). Since a = 1, b = 7, and c = -3, the discriminant is 72 – 4(1)(-3) = 49 + 12 = 61. A positive number under the square root is real, so the answer is (B).


  2. Does the parabola with equation x2 ­– x + 4 = 0 have real or imaginary roots?

    Correct Answer:

    Imaginary, because the discriminant is negative

    Answer Explanation:

    The discriminant of a parabola is b2 – 4ac, which is equal to 1 – 4(1)(4) = -15 in this case. Since the discriminant is under the radical in the quadratic formula, a positive discriminant will yield a real number and a negative will yield an imaginary number. Since the discriminant in this case is -15, the roots of the parabola are imaginary.


  3. Does the parabola with equation -7x2 ­= 18 – 2x have real or imaginary roots?

    Correct Answer:

    Imaginary, because the discriminant is negative

    Answer Explanation:

    In this case, we need to set our equation to equal zero before we can calculate the discriminant. That means we rearrange the equation so that it takes the form -7x2 + 2x – 18 = 0 and take our a, b, and c values from there. If we use those values to calculate the discriminant, we end up with 22 – 4(-7)(-18) = 4 – 504 = -500. The discriminant is very negative, and the square root of negative numbers is why we have imaginary numbers in the first place. That means our answer is (C).


  4. Solve 2x2 + 2x + 5 = 0 for x and express the answer in a + bi form.

    Correct Answer:

    Answer Explanation:

    Since the equation is already set to equal zero, our values are a = 2, b = 2, and c = 5. Plugging that into the quadratic formula, we have , or  That reduces to  or .


  5. Solve x2 + 4x = -6 for x and express the answer in a + bi form.

    Correct Answer:

    Answer Explanation:

    Once we set the equation to equal to zero, a = 1, b = 4, and c = 6. All we need to do is use the quadratic formula and make sure we do the algebra correctly. It will give us , or , which is (B).


  6. Solve 3x2 + 6x = -6 for x and express it in a + bi form.

    Correct Answer:

    -1 ± i

    Answer Explanation:

    If we rearrange the equation to 3x2 + 6x +6 = 0, we can use the values a = 3, b = 6, and c = 6 in quadratic formula. This means we have  or . If we simplify this to its fullest extent, we should end up with (D).


  7. Solve x2 + 2x = -10 for x and express it in a + bi form.

    Correct Answer:

    -1 ± 3i

    Answer Explanation:

    If the equation is rearranged so that a = 1, b = 2, and c = 10, these values can be plugged into the quadratic formula, which gives us , or  In its simplest form, the roots are -1 ± 3i.


  8. Solve  for x and express it in a + bi form.

    Correct Answer:

    Answer Explanation:

    Before we dive headfirst into the quadratic equation, we'll want to rearrange the equation we're given so that it equals zero. Once we do that, we'll have 2x2 + x + 7 = 0. With a = 2, b = 1, and c = 7, we can relish in the beauty that is the quadratic formula, which will give us (C) as the right answer.


  9. Which of the following equations has roots of 3 ± i?

    Correct Answer:

    x2 – 6x + 10

    Answer Explanation:

    Yes, this question means what you think it means. Fortunately, all we need to do is calculate the discriminant of each equation because as it turns out, only one of them is negative. If we calculate them all, we should end up with -36 for (A), 76 for (B) and (D), and 241 for (C). Since the root is imaginary, our only possible answer is (A).


  10. Imaginary roots always come in "conjugate pairs." That means that if the answers are imaginary, we'll get both conjugates as answers. Based on your observations, why do you think it happens?

    Correct Answer:

    Because the only difference between the two answers is the ± sign in the formula

    Answer Explanation:

    The ± sign in the quadratic formula is the reason for the conjugate pairs being the answer. For instance, if a parabola has imaginary roots 3 ± 3i, its roots are actually 3 + 3i and 3 – 3i. The two roots are because of that ± sign in the quadratic formula itself.


Aligned Resources

    More standards from High School: Number and Quantity - The Complex Number System