jazzyfa30
  • jazzyfa30
Solve the equation by using the Square Root Property. x^2 = 10x + 24
Mathematics
katieb
  • katieb
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jazzyfa30
  • jazzyfa30
hartnn
  • hartnn
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
  • hartnn
***the co-efficient of 'x' here is -10

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

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