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**The Discriminant of a Quadratic Equation**

Part 1

Part 1

Recall that the quadratic equation ** ax^{2} + bx + c = 0 **can be solved using the quadratic formula:

For more information on the quadratic formula, see the following posts:

Quadratic Formula – Part 1 Quadratic Formula – Part 2

Before we go on, you may also want to review the information on square roots that can be found here:

The **discriminant** of the quadratic equation ** ax^{2} + bx + c = 0 **is the quantity

**Δ**defined by

**Δ = b^{2} – 4ac**

In other words, the discriminant is simply the expression that appears under the square root in the quadratic formula.

Although computing the discriminant of a quadratic equation does not give the roots (solutions) of the equation, it does give us a lot of information about the nature of the roots and the graph of the equation.

For example, if the discriminant of the quadratic equation ** ax^{2} + bx + c = 0 **is 0, then the quadratic formula simplifies to

** x = –b/2a**,

and we see that there is just one solution.

If the coefficients ** a** and

**are integers, then the unique solution will be a rational number.**

*b*Graphically, this means that the graph of the function * y = ax^{2} + bx + c *is a parabola that intersects the

*x*-axis at one point.

An example is given by the red parabola in the image above.

**Note: **The discriminant **does not** tell us if the parabola opens upwards or downwards. However, this is easy to determine simply by looking at the value of ** a**.

If *a* > 0 (i.e. *a* is a positive number), then the parabola opens upwards.

If *a* < 0 (i.e. *a* is a negative number), then the parabola opens downwards.

So, the red parabola in the image above has an equation of the form *y = ax ^{2} + bx + c* where

*a*< 0 and Δ = 0

**Example: **Find the discriminant of ** x^{2} + 6x + 9 = 0**. Then describe the nature of the roots of the equation, and describe the graph of the function

**y = x**^{2}+**6**

**x +****9**.

I will post a solution to this problem tomorrow, and then discuss other possibilities for the determinant. Feel free to post your own solutions in the comments.

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**Square Root Basics**

A number *a* is a **square root** of a number *x* if *a*^{2} = *x *(or equivalently, *a *⋅ *a* = *x*).

**Example: **3 is a square root of 9 because 3^{2} = 3 ⋅ 3 = 9.

–3 is also a square root of 9 because (–3)^{2} = (–3)(–3) = 9.

We see that 9 has two square roots: 3 and –3. We sometimes combine these two and say that the two square roots of 9 are ±3.

### Square Root Symbol

Students sometimes get confused when using the square root symbol:

The symbol above represents the *positive* square root. So, for example, we have

So even though 9 has the two square roots ±3, when we put 9 under the symbol, the result is just 3 (and not –3). You may want to compare this with the Square Root Property (students often get the square root property confused with taking the positive square root of a number).

If we want the negative square root of a number, we need to place a minus sign before the square root symbol.

And if we want both square roots of a number, we should place the symbol ± before the square root symbol.

### “Types” of Square Roots

All numbers, with the exception of 0, have two square roots. The number 0 has just one square root because both the positive and negative square root of 0 are both 0.

When taking positive and negative square roots of numbers, it’s worth trying to determine the “type” of number that you get. For example, whenever we take the square root of a nonzero integer (the set of integers is {…–3, –2, –1, 0, 1, 2, 3,…}), the result can be an integer, an irrational number, or a pure imaginary number (see this: Complex Numbers). When we identify the type of roots of a number we usually refer to this as *determining the nature of the roots*.

**Examples: **Determine the nature of the roots of 16, 11, –2, and 0.

**Solution:** Since 16 is a positive perfect square, the two square roots of 16 are integers (in fact, they are 4 and –4).

Since 11 is positive, but not a perfect square, the two square roots of 11 are irrational numbers.

Since –2 is negative, the two square roots of –2 are pure imaginary numbers.

Finally, 0 has just one square root…itself. In particular, 0 has one square root, which is an integer.

**Note: **If a positive integer is *not* a perfect square, then it’s square root will always be irrational. The proof of this result is beyond the scope of this article. I will discuss this further in a future post.

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**Derivation of the Quadratic Formula**

Last week we went over how to solve quadratic equations using the quadratic formula. You can see those posts here: Quadratic Formula 1 2

Recall that the solutions to the quadratic equation ** ax^{2} + bx + c = 0 **are

At the end of the second post, I posed the following problem:

**Challenge Problem: **Solve the general quadratic equation *ax*^{2} + *bx *+ *c* = 0 by completing the square, and note that this gives a derivation of the quadratic formula.

I will now provide a detailed solution. You may want to review the following posts before reading the following solution:

Square Root Property Completing the Square Solving Quadratic Equations

**Solution:**

The derivation above was challenging. Try to understand each step. If you have any question on a particular step, feel free to post your question in the comments.

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**The Quadratic Formula**

Part 2

Part 2

Yesterday we began discussing the quadratic formula, and I asked you to solve a specific quadratic equation using the formula. You can see that post here: Solving Quadratic Equations with the Quadratic Formula

Today I will provide the solution to that problem. For your reference, let me give you the quadratic formula one more time:

The solutions to the quadratic equation ** ax^{2} + bx + c = 0 **are

**Example:** Solve the quadratic equation *x*^{2} – 2*x – *15* = *0 by using the quadratic formula.

**Solution:**

So, the two solutions are 1 + 4 = **5**, and 1 – 4 = **–3**

**Note: **This particular problem could be solved more easily by factoring.

Why does the quadratic formula look so messy? It’s not a very pleasant looking formula. It would have been nicer if we had a simpler formula for solving a quadratic equation. But unfortunately, this is simply what it turned out to be. As a challenging exercise, I would like you to see this for yourself firsthand. I will give a full solution tomorrow:

**Challenge Problem: **Solve the general quadratic equation *ax*^{2} + *bx *+ *c* = 0 by completing the square, and note that this gives a derivation of the quadratic formula.

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**Solving Quadratic Equations**

with the Quadratic Formula

with the Quadratic Formula

A **quadratic equation** has the form ** ax^{2} + bx + c = 0**.

The **quadratic formula** is a formula that gives us all solutions to a quadratic equation right away. The formula is as follows:

**Example:** Solve the quadratic equation *x*^{2} – 2*x – *15* = *0 by using the quadratic formula.

We have previously solved this problem by completing the square. You can see that solution here: Solution by Completing the Square

Try to solve this problem by using the quadratic formula. I will give the full solution tomorrow.

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**Complex Numbers**

Solution to Division Problem

Solution to Division Problem

Several days ago I began writing a series of posts about complex numbers. You can see that post here: Complex Numbers Addition and Subtraction Multiplication Division

Today I would like to provide a solution to Thursday’s division problem. Here is the problem one more time, followed by a solution:

**Example:**

**Solution: **

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**Complex Numbers**

Division

Division

A few days ago I began talking about complex numbers, and we learned how to raise the complex number *i* to any power. You can see that post here: Complex Numbers – Examples and Powers of *i*

We then reviewed how to add, subtract, and multiply complex numbers, and I gave you a problem to try involving multiplication. You can see those posts here: Addition and Subtraction Multiplication

Today I will go over how to divide complex numbers. But first I will provide a solution to yesterday’s multiplication problem. Here is the problem one more time, followed by a solution:

**Example: **Compute (2 – 3i)(–5 + 6i)

**Solution: **(2 – 3*i*)(–5 + 6*i*) = (–10 + 18) + (12 + 15)*i* = **8 + 27 i**

**Division**

The **conjugate** of the complex number *a* + *bi* is the complex number *a* – *bi*.

**Examples:**

The conjugate of –5 + 6*i* is –5 – 6*i*.

The conjugate of 3 – 2*i* is 3 + 2*i*.

Note that when we multiply conjugates together we always get a real number. In fact, we have

(*a* + *bi*)(*a* – *bi*) = *a*^{2} + *b*^{2}

We can put the quotient of two complex numbers into standard form by multiplying both the numerator and denominator by the conjugate of the denominator. This is best understood with an example.

**Example:**

**Hint: **Multiply both the numerator and denominator of the fraction by the conjugate of the denominator. The denominator will them become a real number, and the resulting fraction can then be written in the form *a* + *bi*.

I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

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