#1**+1 **

Solving Quadratic Equations by Factoring

1. 3r^2 -16r-7=5

This is a quadratice, (the highest power is 2) so there could be 0,1 or 2 real answers.

**First take it all to one side by subtracting 5 from both sides.**

After you do that I will look at what you have. So unless someone else answers first, the next post is yours.

Melody Feb 12, 2019

#2**0 **

3 r^2 -16r-12 = 0

eeeeek....I had an answer here...but I see Melody is coaching ya thru it.....

ElectricPavlov Feb 12, 2019

#4**+1 **

Anyway, EP did the first bit for you.

Now it looks pretty horrible so I would just use the quadratic formula to solve it.

\(r = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Do you know what a, b and c are?

.Melody Feb 12, 2019

#6**+1 **

When you have a QUADRATIC EQUATION of the form a x^2 + b x + c = 0 (or in this case a r^2 + b r + c = 0)

you can use the QUADRATIC FORMULA to find the values of x (or r in this case)

Melody showed you the FORMULA....Now do you know what a b and c are?

Put those values in the FORMULA and solve it..... OK ?

ElectricPavlov
Feb 12, 2019

#7**+1 **

Ok lets take a look - I'll start from the beginning.

\(3 r^2 -16r-7 = 5\\ \text{you need it to equal 0 so take five from BOTH sides}\\ 3 r^2 -16r-7-5 = 5-5\\ \color{red}{3 r^2 -16r-12 = 0}\\ \text{Now you have it in general from for a quadratic equation}\\ ar^2+br+c=0\\ \color{red}{a=3\qquad b=-16 \qquad c=-12}\\~\\ \text{Now use the quadratic formula to find the answer}\\ r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \)

Now substitute the numbers into the equation and simplify to find your answers. :)

[that was meant to be general FORM not general from ]

Melody Feb 12, 2019

#8**+1 **

3r^2 - 16r - 7 = 5 subtract 5 from both sides

3r^2 - 16r - 12 = 0 we cannot take out any common factor

So....here is a procedure for factoring whenever the lead coefficient is not a 1

1) Multiply the first and last terms = 3 * -12 = -36

2) List all the possible factors of -36 and that add to - 16.....we can stop (if) we find one that adds to the middle term, -16......if we find no combination.....the polynomial is fully factored

-36 1

-18 2

Note that -18 and 2 is what we need

3) Write the original problem in two ways

3x^2 - 18x + 2x - 12 = 0 or

3x^2 + 2x - 18x - 12 = 0

The second can't be factored by grouping, but the first one can

So we have

3x^2 - 18x + 2x - 12 = 0 factor by grouping

3x ( x - 6) + 2 ( x - 6) = 0 the GCF is (x - 6)

( x - 6) ( 3x + 2) = 0 so

x - 6 = 0 or 3x + 2 = 0

x = 6 subtract 2 from both sides

3x = - 2

divide both sides by 3

x = -2/3

The solutions are in red

CPhill Feb 12, 2019