Author Topic: Solving for variable in a quadratic equation (This post is EXTREMELY long)  (Read 571 times)

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Offline Flux3on

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Quadratic Equations, believe it or not, are not that hard. Let's take a look at how to solve one:

Equation: 4x2+4x-3=2

First off, we want to make the 2 in the equation a zero*, so we subtract 2 from both sides:

2-2=0
-3-2=-5

*You don't need to do this if the right number is already zero.

And we'll have the equation we need: 4x2+4x-5=

Now, we assign the numbers to variables of their own using this equation: Ax2+Bx+C. In this case:
A = 4
B = 4
C = -5

To solve for our original variable, x, we need to use the quadratic formula:

x=(-B±√(B2-4AC))/2A

What this means is that there are going to be 2 solutions. Let's solve for X by plugging in A, B, and C:

x=(-4±√(42-4(4)(-5)))/2(4)

I always start by solving what's in the radical.

42-4(4)(-5)
We start with Exponents since there are no multi-term expressions in parentheses.
42=16
Then Multiplication:
-4*4=-16
-16*-5=80
We now have 16+80 which equals 96. We now have that.
x=-4±√(96)/2(4)

Let's multiply the bottom side of the fraction:
2*4=8
x=-4±√(96)/8

For all intents and purposes, I'm converting √(96) to a decimal.
√96 ≈ 9.798

x=-4±9.798/8

We now turn this into 2 equations:
x=-4+9.798/8
x=-4-9.798/8

Then combine terms on the numerator:

x=5.798/8
x=-13.798/8

Then divide.

x=0.72475
x=-1.72475.

So x indeed does have 2 solutions.
x=0.72475,-1.72475

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