Question

Please help me with this math problem. I have to Use the quadratic formula to solve each equation.

Answer 1

You have to solve the given expression for q:

$2q^2-8=3q$

This expression has a quadratic term, which means that it is a quadratic equation. To find the value or values of q, you have to use the quadratic equation, using "q" as the variable instead of "x"

$q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

a is the coefficient of the quadratic term

b is the coefficient of the q term

c is the constant

- First, zero the equation by passing 3q to the left side of the equal sign:

$\begin{gathered} 2q^2-8=3q \\ 2q^2-8-3q=3q-3q \\ 2q^2-3q-8=0 \end{gathered}$

For this equation the coefficients are:

a= 2

b= -3

c= -8

Replace them in the formula and solve:

$\begin{gathered} q=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ q=\frac{-(-3)\pm\sqrt{(-3)^2-4*2*(-8)}}{2*2} \\ q=\frac{3\pm\sqrt{9+64}}{4} \\ q=\frac{3\pm\sqrt{73}}{4} \end{gathered}$

Next, to determine each possible value of q, you have to solve the sum and difference separately:

Sum

$\begin{gathered} q=\frac{3+\sqrt{73}}{4} \\ q=2.886\cong2.9 \end{gathered}$

Difference

$\begin{gathered} q=\frac{3-\sqrt{73}}{4} \\ q=-1.386\cong-1.4 \end{gathered}$

The possible solutions for the given equation are q=2.9 and q=-1.4