### Quadratics - Part 1

What are quadratics? Quadratics are equations where a variable is raised to the second power. For example, take an equation like this:

\(3x^2-6x+9\)

This is an example of quadratic, as the first part is *3**x ^{2}*. The variable is in the second degree, and is this a quadratic. Also on a side note, quadratics functions when graphed make a parabola like this:

Fun Fact: Quadratic is named after "quad" meaning square, because the variable gets squared (like **x ^{2}**). Quadratics are usually in standard form, which looks like this:

\(ax^2-bx+c\)

In the form above, a, b, and c are known values. In the example above, *a* cannot be 0, or it is not quadratic. *x* is the variable we are trying to solve for. Let's try some questions!

### Questions (Standard Form) - Part 1

Find the value for a, b, and c in each question below. Then click on the question to see the answer.

*a = 2*

*b = -4*

*c = 4*

*a = 6*

*b = -1*

*c = -9*

*a = -1*

*b = -3*

*c = 6*

How much did you get? Comment down below!

### Quadratic Formula - Part 2

Now that we have a general idea on how quadratics look like, let's try solving them! There are many different strategies to solve quadratics. In this tutorial, we will be covering the *Quadratic Formula*. The quadratic formula is a formula that is used to solve quadratics (as the name implies). The quadratic formula looks like this:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

It may look like a lot, but it really isn't! In the previous exercise, we found out what a, b, and c was. Using the values, we plug it into the formula above and we will get the value of x! Cool right? Let's try one out!

\(x^2-3x+2=0\)

Let's try solving this! First we need to find out the values of a, b, and c. The coefficient of the first value is 1, so *a=1*. The coefficient of the second value is -3, so *b=-3*. The coefficient of the third value is 2, so c*=2*. Now we can plug it into the formula! Let's try it!

\(x = {-(-3) \pm \sqrt{(-3)^2-4(2)(1)} \over 2(1)}\)

Let's simply things!

\(x = {3 \pm \sqrt{9-8} \over 2}\)

\(x = {3 \pm \sqrt{1} \over 2}\)

\(x = {3 \pm 1 \over 2}\)

Cool! We have simplified this equation. Now we need to solve this. The +/- symbol means add or subtract. This means we have to do both, which means we will have to values of x. Let's try adding them first:

\(x = {3 + 1 \over 2}\)

\(x = {4 \over 2}\)

\(x = 2\)

Nice! So we found the value of x. But there is one more value! Let's now subtract:

\(x = {3- 1 \over 2}\)

\(x = {2 \over 2}\)

\(x = 1\)

Now what is the answer of x? It is 1 **and **2. But how can we check this? A great thing in algebra is we can check everything by plugging back in values from the equation. Let's try plugging in 1 to see if it is true!

\(1^2-3(1)+2=0\)

\(1-3+2=0\)

\(-2+2=0\)

\(0=0\)

What does this provide us? Well does *0=0*. Yes! Therefore, the equation is true, and x does equal 1. If it did not equal 1, than the values would not be equal. We could also check 2 (the other value we found), but I'll leave that to you.

### Questions (Quadratic Formula) - Part 2

Find the value for x in each question below. Then click on the question to see the answer.

x = 5, -7

x = -0.2, -1

x = -3, -4

So that's it guys, I hope you found some value from this post. If I have helped you please be sure to like this post and follow me for part 2. As always, if you have any questions and need help comment down below! This has been your host, Dheirya Tyagi, signing off for another day.

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