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

[bigwill5112]

Homework Statement

C=5.5\sqrt{}x+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?

 

[╔(σ_σ)╝]

Have you tried setting C=R then forming a quadratic and finally using the quadratic formula.

 

[bigwill5112]

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.

 

[╔(σ_σ)╝]

You can square it to get rid of the squareroot. :-)

 

[gomunkul51]

set C=R then square both side, expand.
congratulations you got yourself a quadratic equation.

 

[statdad]

Homework Statement

C=5.5\sqrt{}x+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.
<br /> \begin{align*}<br /> 3.29x &amp; = 5.5 \sqrt{\,x} + 10000 \\<br /> 3.29x - 5.5 \sqrt{\, x} - 10000 &amp; = 0<br /> \end{align*}<br />

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

 

[bigwill5112]

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

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

Ahaa. THANKS! I appreciate everyone's help.