on week seven of precalc we learned how to do quadratic equations with the quadratic formula. The quadratic formula is probably supposed to be the last bastion after trying everything else when it comes to solving equations. It’s like the last thing you’d want to use, while reliable, it takes a minute to set up. It was made to always work and is a lot easier with a calculator as you just plug numbers in. The formula only works problems with the formula Ax²+Bx+C=0 and looks like this

We also watched a song to the tune of pop goes the weasel and that makes it easy remember.

Now, to use the quadratic formula, you need to know where everything is supposed to be. A will replace the x², B is the coefficient to the ‘x’ and C is the remaining constant.

Let’s consider a quadratic equation:

Here, A=2x, B$4$, and C$6$. Using the quadratic formula, we substitute these values:

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So, the solutions to the quadratic equation 2x²+4x−6=0 are x1=1 and $-3$.

In week √36 of precal we learned about quadratic equations. Quadratic equations are unique in the way that they will always be AX^2 +BX + C = 0. An interesting thing about quadratics is that there are different ways of solving them. You can factor them or you can “complete the square”. to complete the square you move the number in the C spot over, find the number that would make it a perfect square, then add it to both sides and simplify further from there. I will go over how to do that, including how to find what number will make it a perfect square.

Lets use the equation X^2+6x+8=0 as an example. To start, we subtract the 8 from the left side and the right side

X^2+6x=-8

Next we find what number goes into the C spot. In this certain equation when there is no coefficient to X^2, it will be half of the coefficient of B squared. In this it will be 3^2 which is 9. We then add 9 to both sides and it will look like this

X^2+6x+9=1

Now we can easily factor the left side to simplify the equation.
(x+3)^2=1

Now we must remove the exponent on the left side and we can do so by rooting both sides

√(x+3)^2=√1 = X+3 = +/-1

because of how rooting works the 1 is either positive or negative, meaning there are 2 answers to these types of equations.

X1=-2    X2=-4

And this is how you solve quadratic equations by “completing the square”.

In week 5 of precal we learned how to factor equations with 4 terms. Using the box method makes doing these things a breeze. Let’s use the equation X^2+6x+2x+12 as an example. we start by drawing a box in the middle and putting the greatest term (usually x^2) at the top left and leave the shortest term (usually a number with no variables) at the bottom right with the terms in the middle (usually variable with no exponent/1) in the other corners. I like to add a positive sign or negative sign on all of the numbers so I don’t make any mistakes.

Now that we have all of our terms in a box, we have to find what goes into all of these terms and place them outside of the box like so:

Now that we have found what goes into the terms we re-write the equation. for this equation it will be (X+6) (X+2) and that is completely factored.

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This week of precal we learned about division of radicals and rationalizing the denominator. It is (obviously) very helpful to rationalize the denominator in a question where the denominator isn’t a rational number. Let’s use this equation as an example:

Now, we know that the denominator is irrational. In order to solve the equation we must rationalize it and the easiest way to do that is by multiplying it by the conjugate of itself (√5+1), remembering that what you do to the bottom you must do to the top.

This is what you get from multiplying the top and bottom by the conjugate of the denominator (√5+1).

In this particular equation we can further simplify it because there is a common factor (4). we can divide the 4s on the top by the 4 on the bottom resulting in √[5+4]. And that is how you simplify a division equation that has an irrational denominator