SOLUTION: I cannot find my error.
I am supposed to use the quadratic formula to solve:
x^2+4x=5
This is what i got:
x=-4+-radical4^2-4*1*5/2*1
-4+-radical 16-20/2
-4+-radical -4/2
Algebra.Com
Question 246960: I cannot find my error.
I am supposed to use the quadratic formula to solve:
x^2+4x=5
This is what i got:
x=-4+-radical4^2-4*1*5/2*1
-4+-radical 16-20/2
-4+-radical -4/2
-4+-2i/2
-2+-i
Hope that makes sense. I am frustrated with this problem
Found 3 solutions by solver91311, checkley77, stanbon:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Your error is that you did not put the quadratic into standard form first.
Standard form is:
But your
is actually in
form.
Either way, your 5 in the quadratic formula has to be -5, thus:
\ \pm\ \sqrt{(4)^2\ -\ 4(1)(-5)}}{2(1)}\ =\ \frac{-(4)\ \pm\ \sqrt{36}}{2}\ =\ \frac{-4\ \pm\ 6}{2}\ =\ -2\ \pm\ 3)
So:
or
Check:
^2\ +\ 4(-5)\ =\ 5)
^2\ +\ 4(1)\ =\ 5)
John
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
x^2+4x=5
x^2+4x-5=0 You forgot to transfer the 5.
You also need to use {( }) for this equation so you won't get lost.
This is what i got:
x=(-4+-radical[4^2-4*1*-5])/2*1
x=(-4+-radical[16+20])/2
x=(-4+-radical36)/2
x=(-4+-6)/2
x=(-4+6)/2
x=2/2=1 ans.
x=(-4-6)/2
x=-10/2
x=-5 ans.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
use the quadratic formula to solve:
x^2+4x=5
Rearrange:
x^2 + 4x -5 = 0
----
x=[-4+-radical(4^2-4*1*-5)]/(2*1)
x = [-4+-radical(16+20)]/2
x = [-4+-radical(36)]/2
x = [-4 +- 6]/2
x = [-10/2] or x = [2/2]
x = -5 or x = 1
=====================
Cheers,
Stan H.
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