Question

Write the following equation in standard form. Then solve.6 q squared minus 5 qequals5 q squared minus 7 q plus 48Question content area bottomPart 1The equation in standard form isenter your response here

Answer 1

To rewrite the equation \( 6q^2 - 5q = 5q^2 - 7q + 48 \) in standard form, we first move all terms to one side of the equation.

Start with the original equation:

\[
6q^2 - 5q = 5q^2 - 7q + 48
\]

Subtract \( 5q^2 \), add \( 7q \), and subtract \( 48 \) from both sides:

\[
6q^2 - 5q - 5q^2 + 7q - 48 = 0
\]

Now, combine like terms:

\[
(6q^2 - 5q^2) + (-5q + 7q) - 48 = 0
\]
\[
q^2 + 2q - 48 = 0
\]

This is the equation in standard form.

Now, we can solve the quadratic equation \( q^2 + 2q - 48 = 0 \) using the quadratic formula:

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

Here, \( a = 1 \), \( b = 2 \), and \( c = -48 \).

First, calculate the discriminant:

\[
b^2 - 4ac = 2^2 - 4(1)(-48) = 4 + 192 = 196
\]

Now apply the quadratic formula:

\[
q = \frac{-2 \pm \sqrt{196}}{2(1)}
\]
\[
q = \frac{-2 \pm 14}{2}
\]

Now, we find the two possible values for \( q \):

1.
\[
q = \frac{-2 + 14}{2} = \frac{12}{2} = 6
\]

2.
\[
q = \frac{-2 - 14}{2} = \frac{-16}{2} = -8
\]

Thus, the solutions of the equation \( q^2 + 2q - 48 = 0 \) are:

**Answer:**

The equation in standard form is \( q^2 + 2q - 48 = 0 \).

The solutions are \( q = 6 \) and \( q = -8 \).