Solve the equation by using the Square Root Property.
x^2 = 10x + 24

- jazzyfa30

Solve the equation by using the Square Root Property.
x^2 = 10x + 24

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- 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

***the co-efficient of 'x' here is -10

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## More answers

- jazzyfa30

so it would be x^2-5x=24

- 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

2.23

- jazzyfa30

is that right????

- hartnn

umm...noo...how u got thta ?

- jazzyfa30

i squared 5 isnt that what i was supposed to do

- hartnn

5^2 =25

- jazzyfa30

oh i thought i was supposed to square root it not square my b now what

- hartnn

ohh...u took square root...we are supposed to square..

- hartnn

ok, we add 25 on both sides
\(x^2-10x+25=24+25\)
notice that left side is a perfect square! of ?

- jazzyfa30

49

- hartnn

49 is right side, and is the square of 7
what about left side ?
x^2-10x+25 is squaree of ?

- jazzyfa30

idk

- hartnn

\((x-5)^2 =... ?\)
can u find ?

- jazzyfa30

x^2- 25

- 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

ok

- jazzyfa30

then what

- jazzyfa30

im confused

- hartnn

did u take square root on both sides ? what u got ?

- jazzyfa30

idk

- 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

but 7 isnt apart of any of my answer choices

- hartnn

u need to solve
x-5 =7
x-5 = -7

- jazzyfa30

x= 12
x=-2

- hartnn

correct!

- jazzyfa30

ok Solve the equation by completing the square: x2 - 8x + 15 = 0

- hartnn

same thing, here, the co-efficient of x is -8
divide it by 2, what u get ?

- jazzyfa30

-4

- hartnn

yes, then square it...what u get ?

- jazzyfa30

-16

- hartnn

-4 *-4 will be +16

- jazzyfa30

oh

- hartnn

so, adding 16 on both sides...
\(x^2-8x+16=-15+16\)

- jazzyfa30

right side:1
left side: idk

- jazzyfa30

i think its x^2-4x-2x=1

- jazzyfa30

is that right??????

- jazzyfa30

@GoldPhenoix please help

- hartnn

\(x^2-8x+16 = (x-4)^2\)
so, we have
\((x-4)^2=1\)
just take square root on both sides!

- jazzyfa30

?????????

- 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

(x-3)(x+3) i think

- jazzyfa30

a. x = + 5
b. x =-3 or x = 5
c. x =3 or x = -5
d. x =3 or x =5

- hartnn

x-4 = -1 , x = -1+4 = 3 is correct
but for 2nd part, you get
x-4 =1
x= 1+4 = 5

- hartnn

so, 3,5

- jazzyfa30

oh yea

- jazzyfa30

Solve the equation by completing the square: x^2 + 2x = 120

- terenzreignz

Do you know how to complete the square? :)

- jazzyfa30

??????

- terenzreignz

I see. Here's an example:\[\Large x^2 + \color{red}6x =40 \]

- terenzreignz

What you do is take that coefficient (number beside) the x with NO EXPONENT. In our case, 6.

- terenzreignz

Take half of it, 3, and then square it, getting 9.

- terenzreignz

Now, add that to BOTH SIDES of the equation.
\[\Large x^2 +\color{red}6x\color{blue}{+9}=40\color{blue}{+9}\]

- terenzreignz

Simplify.
\[\Large x^2+\color{red}6x+9=49\]

- 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

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

So... go ahead now, get started with your own problem, using the same concepts.
I'll walk you through it if you like :)

- jazzyfa30

yes please

- terenzreignz

By your lead :)

- jazzyfa30

u on skype or call me at 803-937-6271 cus by u typing it it is confusing m please

- 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

then call me i dont understand

- jazzyfa30

@jessika96

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