jazzyfa30
Solve the equation by using the Square Root Property.
x^2 = 10x + 24
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jazzyfa30
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@hartnn
hartnn
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ok, so \(x^2-10x=24\)
the co-efficient of 'x' here is 10
divide it by 2 and square the result, what u get ?
hartnn
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***the co-efficient of 'x' here is -10
jazzyfa30
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so it would be x^2-5x=24
hartnn
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no no..
the co-efficient of x is -10, right ?
divide it by 2, you get -5, but then square it, \(\large (-5)^2=...?\)
jazzyfa30
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2.23
jazzyfa30
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is that right????
hartnn
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umm...noo...how u got thta ?
jazzyfa30
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i squared 5 isnt that what i was supposed to do
hartnn
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5^2 =25
jazzyfa30
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oh i thought i was supposed to square root it not square my b now what
hartnn
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ohh...u took square root...we are supposed to square..
hartnn
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ok, we add 25 on both sides
\(x^2-10x+25=24+25\)
notice that left side is a perfect square! of ?
jazzyfa30
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49
hartnn
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49 is right side, and is the square of 7
what about left side ?
x^2-10x+25 is squaree of ?
jazzyfa30
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idk
hartnn
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\((x-5)^2 =... ?\)
can u find ?
jazzyfa30
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x^2- 25
hartnn
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nopes, \((a+b)^2=a^2+2ab+b^2\)
so, \((x-5)^2 =x^2-10x+25\)
so here we have
\((x-5)^2=7^2\)
take square root on both sides!
jazzyfa30
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ok
jazzyfa30
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then what
jazzyfa30
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im confused
hartnn
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did u take square root on both sides ? what u got ?
jazzyfa30
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idk
hartnn
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\((x-5)^2=7^2 \implies x-5 = \pm 7 \\ x-5 =7 \quad \quad x-5=-7\)
get 2 values of 'x' from here!
jazzyfa30
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but 7 isnt apart of any of my answer choices
hartnn
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u need to solve
x-5 =7
x-5 = -7
jazzyfa30
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x= 12
x=-2
hartnn
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correct!
jazzyfa30
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ok Solve the equation by completing the square: x2 - 8x + 15 = 0
hartnn
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same thing, here, the co-efficient of x is -8
divide it by 2, what u get ?
jazzyfa30
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-4
hartnn
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yes, then square it...what u get ?
jazzyfa30
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-16
hartnn
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-4 *-4 will be +16
jazzyfa30
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oh
hartnn
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so, adding 16 on both sides...
\(x^2-8x+16=-15+16\)
jazzyfa30
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right side:1
left side: idk
jazzyfa30
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i think its x^2-4x-2x=1
jazzyfa30
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is that right??????
jazzyfa30
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@GoldPhenoix please help
hartnn
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\(x^2-8x+16 = (x-4)^2\)
so, we have
\((x-4)^2=1\)
just take square root on both sides!
jazzyfa30
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?????????
hartnn
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when you take square root, the square part goes away!
(x+4) =+1 or x-4 =-1
find 2 values of x from here...
jazzyfa30
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(x-3)(x+3) i think
jazzyfa30
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a. x = + 5
b. x =-3 or x = 5
c. x =3 or x = -5
d. x =3 or x =5
hartnn
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x-4 = -1 , x = -1+4 = 3 is correct
but for 2nd part, you get
x-4 =1
x= 1+4 = 5
hartnn
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so, 3,5
jazzyfa30
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oh yea
jazzyfa30
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Solve the equation by completing the square: x^2 + 2x = 120
terenzreignz
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Do you know how to complete the square? :)
jazzyfa30
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??????
terenzreignz
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I see. Here's an example:\[\Large x^2 + \color{red}6x =40 \]
terenzreignz
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What you do is take that coefficient (number beside) the x with NO EXPONENT. In our case, 6.
terenzreignz
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Take half of it, 3, and then square it, getting 9.
terenzreignz
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Now, add that to BOTH SIDES of the equation.
\[\Large x^2 +\color{red}6x\color{blue}{+9}=40\color{blue}{+9}\]
terenzreignz
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Simplify.
\[\Large x^2+\color{red}6x+9=49\]
terenzreignz
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You'll notice that the left side is now what you call a 'perfect square trinomial'
It may be expressed as...
\[\Large (x+\color{red}3)^2=49\]
terenzreignz
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Notice of course, that 3 is half of the original coefficient of x, which was 6.
Now take the square root of both sides:
\[\Large \sqrt{(x+3)^2}=\sqrt{49}\]
Notice that the square root and the square (exponent 2) cancel out.
But the square root of 49 may be either positive or negative 7.
\[\Large x+3=\pm7\]
Bring the 3 to the other side, you now get
\[\Large x = -3\pm7\]
which means either
x = -3 + 7 = 4
or
x = -3 - 7 = -10
Case Closed :)
terenzreignz
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So... go ahead now, get started with your own problem, using the same concepts.
I'll walk you through it if you like :)
jazzyfa30
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yes please
terenzreignz
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By your lead :)
jazzyfa30
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u on skype or call me at 803-937-6271 cus by u typing it it is confusing m please
terenzreignz
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I don't have skype.
come on, we can do this... I'll get you started...
\[\Large x^2 +\color{red}2x = 120\]
jazzyfa30
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then call me i dont understand
jazzyfa30
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@jessika96