SOLUTION: Hi, I am currently trying to solve this equation. (x+4)^2=3/25 The first thing I tried was this (x+4)(x+4)=3/25 x^2+8x+16=3/25 25*x^2+8x+16=3/25*25 25x^2+120x+340=3 25x^2

Algebra ->  Radicals -> SOLUTION: Hi, I am currently trying to solve this equation. (x+4)^2=3/25 The first thing I tried was this (x+4)(x+4)=3/25 x^2+8x+16=3/25 25*x^2+8x+16=3/25*25 25x^2+120x+340=3 25x^2      Log On


   

Question 692113: Hi, I am currently trying to solve this equation.
(x+4)^2=3/25
The first thing I tried was this
(x+4)(x+4)=3/25
x^2+8x+16=3/25
25*x^2+8x+16=3/25*25
25x^2+120x+340=3
25x^2+120x+340-3=0
25x^2+120x+337=0
But now I can't factor that so I seem to have made a mistake somewhere. Can anyone help? Thx in advance :)

and this is where I am sure I've made some huge mistake. Can anyone help me? Thx so much.
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
Take the square root of both sides
%28x%2B4%29%5E2=3%2F25
...
sqrt%28x%2B4%29%5E2 = +/-sqrt%283%2F25%29
x+4 = +/- sqrt%283%29%2F5
highlight_green%28x+=+-4+%2B-sqrt%283%29%2F5%29
...
Two Solutions
x = -4%2Bsqrt%283%29%2F5
x = -4-sqrt%283%29%2F5
...
Alternatively,
(x+4)(x+4)=3%2F25
x%5E2+%2B+8x+%2B+16+=+3%2F25 ==> multiply by 25
25(x%5E2+%2B+8x+%2B+16+=+3%2F25)
25x%5E2+%2B+200x+%2B+400+=+3
25x%5E2+%2B+200x+%2B+397+-+0
Apply the quadratic formula
x = %28-200%2B-sqrt%28200%5E2-4%2825%29%28397%29%29%29%2F%282%2A25%29
x = %28-200%2B-sqrt%2840000-39700%29%29%2F50
x = %28-200%2B-sqrt%28300%29%29%2F50
x = %28-200%2B-10sqrt%283%29%29%2F50
x = -4%2B-sqrt%283%29%2F5
...
In short, you erred within the multiplication of the equation by 25
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