We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#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