SOLUTION: Solve 25x^2 + 49 = 0 This is what I have: 25x^2 + 49 - 49 = 0 + -49 25x^2 = -49 25x^2/25 = -49/25 x^2 = -1.96 x^2 + 1.96 = 1.96 + - 1.96 x^2 + 1.96 = 0 Therefore a soluti

Algebra ->  Equations -> SOLUTION: Solve 25x^2 + 49 = 0 This is what I have: 25x^2 + 49 - 49 = 0 + -49 25x^2 = -49 25x^2/25 = -49/25 x^2 = -1.96 x^2 + 1.96 = 1.96 + - 1.96 x^2 + 1.96 = 0 Therefore a soluti      Log On


   

Question 280547: Solve 25x^2 + 49 = 0
This is what I have:
25x^2 + 49 - 49 = 0 + -49
25x^2 = -49
25x^2/25 = -49/25
x^2 = -1.96
x^2 + 1.96 = 1.96 + - 1.96
x^2 + 1.96 = 0
Therefore a solution cannot be determine. Am I correct or is the solution x^2 + 1.96 = 0?
Found 2 solutions by Alan3354, Earlsdon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
1.96 = 1.4^2
x^2 = -1.96 (a few steps back)
x = ± 1.4i
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There are no solutions using real numbers.
i = sqrt(-1)

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
25x%5E2%2B49+=+0
You're on the right track!
25x%5E2%2B49+=+0 Subtract 49 from both sides.
25x%5E2+=+-49 Take the square root of both sides.
sqrt%2825x%5E2%29+=+sqrt%28-49%29
5x+=+7sqrt%28-1%29 or 5x+=+-7sqrt%28-1%29 Divide both sides by 5.
x+=+%287%2F5%29sqrt%28-1%29 or x+=+%28-7%2F5%29sqrt%28-1%29 which can also be expressed as:
x+=+%287%2F5%29i or x+=+-%287%2F5%29i where: i+=+sqrt%28-1%29
There are no real solutions.