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

Feb 12, 2019

#1
+107480
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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.

Feb 12, 2019
#2
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3 r^2 -16r-12 = 0

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

Feb 12, 2019
#3
+107480
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Thanks EP.

Nerd123 asked me privately to help him. I've told him I will coach him through it.

I will stay here a little while to give him a chance to respond. :)

Melody  Feb 12, 2019
#4
+107480
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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?

.
Feb 12, 2019
edited by Melody  Feb 12, 2019
#5
+299
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No, can you please explain to me?

Nerd123  Feb 12, 2019
#6
+20217
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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
+107480
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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 ]

Feb 12, 2019
edited by Melody  Feb 12, 2019
#8
+107348
+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

Feb 12, 2019