i have x=plus minus the square root of 119 divided by 8
2006-11-19 06:57:42 · 6 answers · asked by Charnissia S 1 in Science & Mathematics ➔ Mathematics
The answer is +/- the square root of 119, over 8. Divide both sides of the equation by 8, and you get X squared = 119/64. Take the square root of this and you have the answer. 64 is a perfect square, but 119 isn't. If it had been 121 on top, your answer would be +/- 11/8.
2006-11-19 07:05:43 · answer #1 · answered by TitoBob 7 · 0⤊ 0⤋
Well lets see if I can help you here, when you say you have x = + or - the sqr I think of the quadratic equation ;
x = -b + or - sqr(b^2 -4ac)/2a which can only be worked if your original problem is in the form ax^2 + bx + c = 0. To do that we ahve to work the equation a little bit.
1: multiply both sides by 8 and rewrite the equation 64x^2 = -119
2: now add 119 to both sides and rewrite 64x^2 + 119 = 0
3: insert the missing term 64x^2 + x +119 = 0
4: plug you variables into the quadratice equation and work the problem.
x = -b + sqr(b^2 - 4ac)/2a and x = -b - sqr(b^2 - 4ac)/2a and just incase you are corn fused a = 64 b = 1 c = 119.
As always "Please Excuse My Dear Aunt Sally"
2006-11-19 08:09:57 · answer #2 · answered by ikeman32 6 · 0⤊ 0⤋
The answer is incorrect. You cannot take the square root of a negative number unless you incorporate imaginary numbers. Therefore, if I understand your equation correctly 8x^2 = -119/8 then the answer is x = sqrt[119/64]i where i is the imaginary number denoted by the sqrt of -1.
2006-11-19 07:05:44 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The correct answer is ± sqrt(-119) / 8.
Don't forget it's minus 119.
2006-11-19 07:07:18 · answer #4 · answered by falzoon 7 · 0⤊ 0⤋
You have the correct answer.
2006-11-19 07:00:11 · answer #5 · answered by ignoramus 7 · 0⤊ 0⤋
yay good job!! u got it!
2006-11-19 07:04:48 · answer #6 · answered by yook 2 · 0⤊ 0⤋