Question

question 7 | 3 - 8 solving linear and non - linear systems
a small business owner determines that the profit p in dollars from sales of a specific item can be modeled by p = 2x²+30x, where x is the selling price of the item in dollars.
create a linear - quadratic system to determine the price for which the business will earn $50,000.
solve the system in part a to determine the price for which the business will earn $50,000. round to the nearest hundredth, if necessary.
what is the solution set of the system mean in the context of the situation?

Explanation:

Step1: Set up the profit - equation

We know that $P = 50000$ and $P=2x^{2}+30x$. So, we set up the quadratic equation $2x^{2}+30x - 50000=0$. Divide the entire equation by 2 to simplify: $x^{2}+15x - 25000 = 0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $x^{2}+15x - 25000=0$, we have $a = 1$, $b = 15$, and $c=-25000$. First, calculate the discriminant $\Delta=b^{2}-4ac=(15)^{2}-4\times1\times(-25000)=225 + 100000=100225$.

Step3: Find the values of x

$x=\frac{-15\pm\sqrt{100225}}{2}=\frac{-15\pm316.58}{2}$. We have two solutions: $x_1=\frac{-15 + 316.58}{2}=\frac{301.58}{2}=150.79$ and $x_2=\frac{-15 - 316.58}{2}=\frac{-331.58}{2}=-165.79$. Since the selling - price $x$ cannot be negative, we discard the negative solution.

Answer:

$x\approx150.79$