solve using quadratic formula with steps x^2-x-30=0
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OpenStudy (anonymous):
I can solve it without the quadratic formula
OpenStudy (brittany):
oh how
OpenStudy (anonymous):
Well, there's three ways you can solve it without the quad formula:
1. Graph It
2. Complete The Square
3. Factor By Grouping
OpenStudy (brittany):
no thanks i need to learn quadratic formula
OpenStudy (anonymous):
Graphing is the easiest of course, but it requires a graphing calc
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OpenStudy (anonymous):
Well, when using the quadratic formula, you always have to identify a, b, and c. Do you know how to do that?
OpenStudy (brittany):
yes
OpenStudy (brittany):
a=x^2 b=-x c= -30
OpenStudy (anonymous):
Good, that will help you
OpenStudy (anonymous):
Now, the only thing we have to do is know the quadratic formula and input the values of a, b, and c into it, then simplify
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OpenStudy (brittany):
ok i usually have trouble simplfing it because idk wht step is first
OpenStudy (anonymous):
\[\frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
OpenStudy (anonymous):
I see, well, we will plug in the values of a,b, and c first. You always use order of operations when simplifying expressions. The square root is the same as an exponent, so you have to evaluate that first.
OpenStudy (brittany):
in the square root should be x^2-4(x^2)(-30)
OpenStudy (anonymous):
Oh wait....your a, b, and c are incorrect
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OpenStudy (anonymous):
a, b, and c, are numbers only that come before the variables in each term
OpenStudy (brittany):
oh?
OpenStudy (brittany):
x^2-x-30=0
a=1 b= -1 c = -30
OpenStudy (anonymous):
excellent
OpenStudy (anonymous):
Now what goes inside the square root?
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OpenStudy (brittany):
-1^2-4(1)(-30)
OpenStudy (anonymous):
(-1)^2 -4(1)(-30)
Good but make sure you put the -1 inside parentheses
OpenStudy (anonymous):
Because you will be squaring -1
OpenStudy (brittany):
which would make it 1
OpenStudy (anonymous):
Yes
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OpenStudy (brittany):
1-4(1)(-30)
OpenStudy (anonymous):
excellent
OpenStudy (brittany):
now what
OpenStudy (anonymous):
Now you finish simplifying everything inside the square root
OpenStudy (brittany):
how?
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OpenStudy (anonymous):
Order of operations still apply
OpenStudy (anonymous):
You multiply first before subtracting
OpenStudy (brittany):
multiply (1)(-30)?
OpenStudy (anonymous):
Multiply 4*1*-30
OpenStudy (brittany):
-120?
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OpenStudy (anonymous):
yes, now you have 1-(-120)
OpenStudy (brittany):
121
OpenStudy (anonymous):
very good. Do you know how to take the square root of that?
OpenStudy (brittany):
well i normally wld use a calculater but i dont have one
OpenStudy (anonymous):
What is 11*11?
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OpenStudy (brittany):
121
OpenStudy (anonymous):
Yes, so the square root of 121 is 11
OpenStudy (brittany):
ok so now its 1+-11/?
OpenStudy (anonymous):
\[\frac{1 \pm 11}{2*1}\]
OpenStudy (brittany):
1±11/2
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OpenStudy (anonymous):
well, actually if you put it in that form, it isn't correct
OpenStudy (brittany):
oh?
OpenStudy (anonymous):
\[(1\pm 11)/2\]
OpenStudy (anonymous):
You have to make sure you put parentheses, that way we know that the whole numerator is divided by 2, not just 11/2
OpenStudy (brittany):
ok
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OpenStudy (anonymous):
It's what happens when you use the linear form
OpenStudy (brittany):
how do i diivode tht by two
OpenStudy (anonymous):
If you write it as a top down fraction, you can leave it as:
\[x = \frac{1 \pm 11}{2}\]
OpenStudy (anonymous):
Usually, you just leave it in that form because dividing by two would give decimals instead of whole numbers. We started with whole numbers so we should leave the answer in whole numbers
OpenStudy (brittany):
if i am aloud to use deciamls what do i do
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OpenStudy (brittany):
i canuse deciamls rounded to the nearest tenth
OpenStudy (anonymous):
Then you can write it as:
\[x = .5 \pm 5.5 \]
OpenStudy (brittany):
there positive?
OpenStudy (anonymous):
Or you can split it up into two values of x:
x = -5
x = 6
OpenStudy (anonymous):
So in fact, we do end up with whole numbers, so you can put that as the answer
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