SOLUTION: Use the quadratic formula to solve the following. Leave irrational roots in the simplest radical form.
2x^2-3x+1=0
ive got so far the last result x=3_+√3*5
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Question 763958: Use the quadratic formula to solve the following. Leave irrational roots in the simplest radical form.
2x^2-3x+1=0
ive got so far the last result x=3_+√3*5
------------
4
and then im stuck and confuse.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Use the quadratic formula to solve the following. Leave irrational roots in the simplest radical form.
2x^2-3x+1=0
----
x = [3 +- sqrt(9-4*2*1)]/4
---
x = [3 +- sqrt(1)]/4
----
x = (3+1)/4 or x = (3-1)/4
----
x = 1 or x = 1/2
====================
Cheers,
Stan H.
=============
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Not sure how you got to 3 times 5 under the radical. The number under the radical is (1)\ =\ 9\ -\ 8\ =\ 1)
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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