Solve Break-Even Question: C=5.5\sqrt{}x+10,000, R=3.29x

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Homework Statement


C=5.5[tex]\sqrt{}x[/tex]+10,000
R=3.29x
C=R

Homework Equations





The Attempt at a Solution


I really don't have an attempt. Accountant tryin to learn calculus from his old book. I don't know where to start or I guess i should say i don't know how to factor. The book usually shows an emblem of a graphing calculator when you should use one and this problem doesn't have the emblem. so how do I solve without calculator? Thanks. Been a part of many forums, but never a math forum. Am i posting in the correct area?
 
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Have you tried setting C=R then forming a quadratic and finally using the quadratic formula.
 
I have. However, there is an X to the 1/2. Will the quadratic still work? Sorry if this is really easy. Just gettin back ino it.
 
You can square it to get rid of the squareroot. :-)
 
set C=R then square both side, expand.
congratulations you got yourself a quadratic equation.
 
bigwill5112 said:

Homework Statement


C=5.5[tex]\sqrt{}x[/tex]+10,000
R=3.29x
C=R

Homework Equations





The Attempt at a Solution


I really don't have an attempt. Accountant tryin to learn calculus from his old book. I don't know where to start or I guess i should say i don't know how to factor. The book usually shows an emblem of a graphing calculator when you should use one and this problem doesn't have the emblem. so how do I solve without calculator? Thanks. Been a part of many forums, but never a math forum. Am i posting in the correct area?

You don't have to square a thing to solve this.
[tex] \begin{align*}<br /> 3.29x & = 5.5 \sqrt{\,x} + 10000 \\<br /> 3.29x - 5.5 \sqrt{\, x} - 10000 & = 0<br /> \end{align*}[/tex]

Let [tex]u = \sqrt x[/tex] and substitute - you get a quadratic to solve, then you can back-substitute to find [tex]x[/tex]
 
statdad said:
You don't have to square a thing to solve this.
[tex] \begin{align*}<br /> 3.29x & = 5.5 \sqrt{\,x} + 10000 \\<br /> 3.29x - 5.5 \sqrt{\, x} - 10000 & = 0<br /> \end{align*}[/tex]

Let [tex]u = \sqrt x[/tex] and substitute - you get a quadratic to solve, then you can back-substitute to find [tex]x[/tex]

Ahaa. THANKS! I appreciate everyone's help.