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Solving Quadratic Equations by Factoring

 

1. 3r^2 -16r-7=5 

 Feb 12, 2019
 #1
avatar+100819 
+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.

 Feb 12, 2019
 #2
avatar+18360 
0

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
avatar+100819 
+1

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
avatar+100819 
+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?

.
 Feb 12, 2019
edited by Melody  Feb 12, 2019
 #5
avatar+307 
+1

No, can you please explain to me?

Nerd123  Feb 12, 2019
 #6
avatar+18360 
+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 ?   cheeky

ElectricPavlov  Feb 12, 2019
 #7
avatar+100819 
+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 ]

 Feb 12, 2019
edited by Melody  Feb 12, 2019
 #8
avatar+100570 
+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

 

 

cool cool cool

 Feb 12, 2019

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