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.Com
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) (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
--------------
There are no solutions using real numbers.
i = sqrt(-1)
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Solve:
You're on the right track!
Subtract 49 from both sides.
Take the square root of both sides.
or Divide both sides by 5.
or which can also be expressed as:
or where:
There are no real solutions.
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