Question

What are the dimensions of the following rectangle?
3x+2
x A=120 x
3x+2

A. -6 2/3, -18
B. -6 2/3, 6
C. 6
D. 6, 20

Answer 1

@johnweldon1993 can you help me??

Answer 2

@formerlyadinosaur can you help me??

Answer 3

Wait hang on...so does the rectangle look like

Answer 4

Yep, that's what it looks like. :-)

Answer 5

Oh okay....

Alright well first...what do we know about the area of a triangle?

\[\large Area = Length \times Width\] correct?

Answer 6

Area of a rectangle!! **** lol typo there

Answer 7

Yep, :-)

Answer 8

Alright....so we know that the length here will be 3x + 2 and the width would be 'x'....so we just plug those into that equation

\[\large Area = (3x + 2) \times x\]

Answer 9

Now we know the total Area would be the 120 they gave....so we can plug that in too...now we have

\[\large 120 = (3x + 2) \times x\] good so far?

Answer 10

Yep, so far. :-)

Answer 11

Alright now lets simplify the right side there....

\[\large (3x + 2) \times x\]

If we distribute that 'x' into both terms...we will have

\[\large 3x^2 + 2x\] right?

Answer 12

That's what I got.

Answer 13

Okay...so we have

\[\large 3x^2 + 2x = 120\]

Well this just looks like a quadratic here...which can either be solved by the quadratic formula...or completing the square...which would you prefer?

Answer 14

The quadratic.

Answer 15

Alright...so first...lets rewrite our equation as

\[\large 3x^2 + 2x - 120 = 0\]

okay?

Answer 16

Yep, :-)

Answer 17

So now the quadratic formula looks like

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

look familiar?

Answer 18

Yep it does. :-)

Answer 19

In that formula...the 'a' 'b' and 'c' represent the coefficients of the equation we have...

\[\large \color \red{3}x^2 + \color \red{2}x - \color \red{120}\]

so here....

\[\large a = 3\]
\[\large b = 2\]
\[\large c = -120\]

so lets plug those into the quadratic formula

Answer 20

\[\large \frac{-2 \pm \sqrt{2^2 - 4(3)(-120)}}{2(3)}\]

Can you simplify that down for me?

Answer 21

OK,

Answer 22

Yes, just a min please.

Answer 23

Ever figure this out?

Answer 24

I got \[-25/3\]

Answer 25

\[\large \frac{-2 \pm \sqrt{2^2 - 4(3)(-120)}}{2(3)}\]
\[\large \frac{-2 \pm \sqrt{2^2 + 1440}}{2(3)}\]
\[\large \frac{-2 \pm \sqrt{1444}}{2(3)}\]
\[\large \frac{-2 \pm 38}{6}\]

so we have 2 equations...but 1 solution is negative so we will ignore it...

\[\large \frac{36}{6} = 6\]

This solved for our 'x'...now we just plug that in fro 'x' in our length and width

\[\large length = 3x + 2 \rightarrow 3(6) + 2 \rightarrow 18 + 2 \rightarrow 20\]
And
\[\large width = x = 6\]

So if that is true...we know length = 20 and width = 6

Answer 26

Thanks a bunch!!! You are really smart If you ever need help with a question I will try my best at it!!! Bye!! ;-)

Answer 27

No problem :)