i need to solve this radical equation, so far i've gotten this far,
x=sqrt3x+28
then.....
x^2=3x+28
how do i solve the next, please help, help is appreciated.!!!
2006-12-06 14:31:40 · 11 answers · asked by BLUEEEE 1 in Science & Mathematics ➔ Mathematics
x^2 - 3x - 28 = 0
(x-7)(x+4) = 0
x = 7 or x = -4
2006-12-06 14:36:40 · answer #1 · answered by martina_ie 3 · 0⤊ 0⤋
answer is 7
set = to 0 and solve as such:
x^2 -3x -28 =0 I would use factor method since I see 7 and 4 are factors of 28 that have a difference of 3
(x - 7) (x +4) = 0 so
x - 7 = 0 or x + 4 = 0
x = 7 or x = -4 go back to orginal equation and you'll see that x can't be negative because it's thr result of a square root - you can also plug in -4 and see that it won't work since -4 doesn't = the square root of 3*(-4) +28.
However, when you plug in 7, you get a true equation:
7= sqrt 3*7 -+28
7 = sqrt of 21+28
7 = sqrt of 49
7 = 7
:-)
2006-12-06 14:38:54 · answer #2 · answered by NvestR3322 2 · 0⤊ 0⤋
x^2=3x+28
x^2-3x-28=0 Set it up equal to zero
(x-7)(x+4)=0 Factor the polynomial
Zeros are
x=7 and x=-4
2006-12-06 14:39:12 · answer #3 · answered by Loves iT 2 · 0⤊ 0⤋
For one thing, if your equation is indeed
x = sqrt(3x) + 28
Then you have to move the 28 to the left hand side, to get
x - 28 = sqrt(3x)
At this point, when you square both sides, you get
(x - 28)^2 = 3x
x^2 - 56x + 784 = 3x
x^2 - 59x + 784 = 0
Are you sure you wrote your question down right? The answer isn't pretty.
2006-12-06 14:35:27 · answer #4 · answered by Puggy 7 · 0⤊ 0⤋
subtract x^2from both sides
0=-x^2+3x+28
find the factors, using quadratic equation or what ever.
(x+7) and (x-4)
so -7 and 4 are the 2 solutions.
i would work out the quadratic, but my computer does not make those symbols and you should be able to find it or know it. just plug the numbers in.
2006-12-06 14:38:55 · answer #5 · answered by jabber_wok 2 · 0⤊ 0⤋
x=sqr(3x) + 28
Change it a little:
sqr(3x) =28-x
now square both sides
3x = 28^2 - 56x + x^2 and make it into a quadratic
x^2 - 59x + 784 = 0
Apply the quadratic formula and you are in
2006-12-06 14:41:23 · answer #6 · answered by kellenraid 6 · 0⤊ 0⤋
x^2-3x-28=0
(x-7)(x+4)
x=7,x=-4
2006-12-06 14:35:06 · answer #7 · answered by Anonymous · 0⤊ 0⤋
x^2-3x-28=0
x^2-7x+4x-28=0
x(x-7)+4(x-7)=0
(x-7)(x+4)=0
x-7=0 so x=7
x+4=0 sox=-4
so x=-4 or 7
2006-12-06 14:36:06 · answer #8 · answered by raj 7 · 0⤊ 0⤋
Do you mean x = sqrt(3x+28) or x = sqrt(3x) + 28?
Either way you're all doing it wrong.
2006-12-06 14:35:56 · answer #9 · answered by syphonbyte 2 · 0⤊ 0⤋
you make it into a quadratic equation:
x^2-3x-28=0
and you get
x= 7
and
x= -4
the end.
2006-12-06 14:34:55 · answer #10 · answered by brookbabe90 5 · 0⤊ 0⤋