Solve the equation. Express the solution as a radical in the simplest form.

- anonymous

Solve the equation. Express the solution as a radical in the simplest form.

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- katieb

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- jhonyy9

where is ?

- anonymous

|dw:1314895631260:dw|

- jhonyy9

16(x+3)=x2
x2 -16x -48=0

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## More answers

- anonymous

do you mean x^2?

- anonymous

square both sides.

- jhonyy9

yes

- anonymous

oh ok

- anonymous

Just make sure that whatever solutions you find are valid for the original question.

- anonymous

ok

- anonymous

(x-12)(x-4)

- anonymous

how did you get that ?

- anonymous

I have no idea, let me try it again

- anonymous

It's :
x^2-16x-48=0
not:
x^2-16x+48=0

- anonymous

ok

- anonymous

I got tht

- anonymous

did I get the answer right though?

- anonymous

You got this right:
x^2-16x+48 = (x-12)(x-4)
But we need to solve this:
x^2-16x-48=0
Probably easiest to just use the quadratic formula.

- anonymous

ok i'll try that

- anonymous

24,-40

- anonymous

can you show how you got that ?

- anonymous

\[ax^2+bx+c=0\]
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
What values did you use for:
a=?
b=?
c=?

- anonymous

sorry my comp was dissconnected again

- anonymous

ok - so are you ready to continue solving the above ?

- anonymous

yeah, I got -4,-10

- anonymous

what values did you use for
a=?
b=?
c=?
when using the quadratic formula ?

- anonymous

a=1 b=-16 c=-48

- anonymous

right. can you show your full calculation ?

- anonymous

first, can you calculate this part ?
b^2-4*a*c
That's the part inside the square root.

- anonymous

|dw:1314900092634:dw||dw:1314899842897:dw|

- anonymous

the top drawing is supposed to be on bottom

- anonymous

I'll start with this guy under the square root. He is a special guy and even has a name: The discriminant.
\[\Delta=b^2-4ac\]
\[\Delta=b^2-4ac=(-16)^2-4(1)(-48)=256+192=448\]
so,
\[x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\]
\[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]

- anonymous

let me know if the above is clear.

- anonymous

i'm trying to understand it

- anonymous

I don't get it

- anonymous

can you tell me where it becomes unclear ?

- anonymous

you basically did the same thing in the drawings you posted -
just the negative/positive signs got messed up a bit.

- anonymous

oh I get it, thanks

- anonymous

we're not done yet though ... so if anything above is unclear, let me know before we go on.
:)

- anonymous

ok, explain the third line

- anonymous

do you mean the part where the number 8 suddenly pops out from under the square root ?

- anonymous

7*8^2

- anonymous

oh - this ?
448 = 7*64 = 7*8*8 = 7*8^2

- anonymous

yeah

- anonymous

the reason I wrote it that way, is because the question wants us to:
"Express the solution as a radical in the simplest form"
Which means that we need to take everything we can out from under the square root.
\[\sqrt{448}=\sqrt{7*64}=\sqrt{7*8^2}=\sqrt{7}\sqrt{8^2}=\sqrt{7}*8=8\sqrt{7}\]
Is that cleat now ?

- anonymous

yeah

- anonymous

ok but we're still not done ...

- anonymous

The quadratic equation gave us two solutions for x:
\[x=8+4\sqrt{7}\]
or
\[x=8-4\sqrt{7}\]
Right ?

- anonymous

yeah

- anonymous

but if you look at the original question, which was:
\[4\sqrt{x+3}=x\]
Notice the there is a square root there, and (x+3) is inside the square root.
Right ?

- anonymous

yeah

- anonymous

Can a negative number have a real square root ?

- anonymous

no

- anonymous

so that means that since (x+3) is inside the square root,
we cannot allow (x+3) to be negative - right ?

- anonymous

right

- anonymous

so we need to check the two possible values we got for x from the quadratic equation, and check that they don't make the number (x+3) a negative number.
One of the numbers is easy to check, try checking this one first:
\[x=8+4\sqrt{7}\]
When you plug the above value into:
\[(x+3)\]
Do you get a negative value, or not ?

- anonymous

positive

- anonymous

is tht the answer?

- anonymous

yes it is.

- anonymous

ok thanks :)

- anonymous

\[16(x + 3) = x^2\]
\[x^2 - 16x - 48= 0 \]

- anonymous

I did squaring ....

- anonymous

Now I will find Solutions

- anonymous

\[x = 4(2 +\sqrt7)\]
\[x = 4(2 - \sqrt7)\]
http://www.wolframalpha.com/input/?i=x^2+-+16x+-+48+%3D+0

- anonymous

Now when we put the solution in Original function

- anonymous

\[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 -\sqrt7)\)

- anonymous

You will get - in root which is not possible that is why only one solution is possible

- anonymous

yes,
but plugging in that value into (x+3), it is not negative:
(x+3) = (4*(2-sqrt(7)))+3 = 0.41...

- anonymous

Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong

- anonymous

Let me check wolfram then

- anonymous

http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8+-+4sqrt7%29%29%3D8+-+4sqrt7

- anonymous

Check this out

- anonymous

Checking: 8+4*sqrt(7) - Wolfram says true:
http://www.wolframalpha.com/input/?i=4*sqrt%28%288%2B4*sqrt%287%29%29%2B3%29%3D%288%2B4*sqrt%287%29%29
Checking: 8-4*sqrt(7) - Wolfram says false:
http://www.wolframalpha.com/input/?i=4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

- anonymous

Yup

- anonymous

Something wrong with Wolfram I guess

- anonymous

hhhmmmm...
http://www.wolframalpha.com/input/?i=-4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

- anonymous

A result of squaring both sides ...

- anonymous

wolfram is right.

- anonymous

http://www.wolframalpha.com/input/?i=sqrt%2811+-4sqrt7%29*4
yeah you got it squaring removes the minus sign

- anonymous

Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..

- anonymous

yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.

- anonymous

coz only one of them is valid.

- anonymous

well I don't think teacher would expect that ...hmm maybe they allow calculators

- anonymous

ok -- we can ask wolfgirl what they usually do in class.
Maybe she's related to wolfram.

- anonymous

Lol ...
Good Job though ...really nice

- anonymous

I'm not related to wolfram lol. But so far I am clueless as to what they do in class. I do Ashworth online school if that helps any

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