College

Solve the quadratic equation by completing the square.

[tex]3f^2 + 84f - 141 = 0[/tex]

Write your answers as integers, proper or improper fractions in simplest form, rounded to the nearest hundredth.

[tex]f = \square \text{ or } f = \square[/tex]

Answer :

We start with the quadratic equation

[tex]$$
3f^2 + 84f - 141 = 0.
$$[/tex]

Step 1. Divide by 3

Divide every term by 3 to simplify:

[tex]$$
\frac{3f^2}{3} + \frac{84f}{3} - \frac{141}{3} = 0 \quad \Longrightarrow \quad f^2 + 28f - 47 = 0.
$$[/tex]

Step 2. Isolate the quadratic and linear terms

Move the constant to the right side:

[tex]$$
f^2 + 28f = 47.
$$[/tex]

Step 3. Complete the square

To complete the square, take half of the coefficient of [tex]$f$[/tex], which is [tex]$\frac{28}{2} = 14$[/tex], and square it:

[tex]$$
14^2 = 196.
$$[/tex]

Add [tex]$196$[/tex] to both sides:

[tex]$$
f^2 + 28f + 196 = 47 + 196,
$$[/tex]

which simplifies to

[tex]$$
f^2 + 28f + 196 = 243.
$$[/tex]

The left side is now a perfect square:

[tex]$$
(f + 14)^2 = 243.
$$[/tex]

Step 4. Solve for [tex]$f$[/tex]

Take the square root of both sides:

[tex]$$
f + 14 = \pm\sqrt{243}.
$$[/tex]

Since

[tex]$$
\sqrt{243} \approx 15.588457268119896,
$$[/tex]

solve for [tex]$f$[/tex] by subtracting [tex]$14$[/tex]:

[tex]$$
f = -14 \pm 15.588457268119896.
$$[/tex]

This gives two solutions:

1.

[tex]$$
f = -14 + 15.588457268119896 \approx 1.588457268119896 \approx 1.59,
$$[/tex]

2.

[tex]$$
f = -14 - 15.588457268119896 \approx -29.588457268119896 \approx -29.59.
$$[/tex]

Final Answer:

[tex]$$
f \approx 1.59 \quad \text{or} \quad f \approx -29.59.
$$[/tex]