How do you solve #3x^2 - 8x + 5 = 0 # using the quadratic formula?

1 Answer
Aug 12, 2016

The two possible answers are: #x = 1.667# and #x=1#

Explanation:

I'll provide the quadratic formula so you can see what I'm doing as I step you through the process
mswilsonalgebra1.weebly.com

I think it's worthwhile to mention that #a# is the number that has the #x^2# term associated with it. Thus, it would be #3x^(2)# for this question.#b# is the number that has the #x# variable associated with it and it would be #-8x#, and #c# is a number by itself and in this case it is 5.

Now, we just plug our values into the equation like this:

#x = (- (-8) +- sqrt((-8)^(2) - 4(3)(5)))/(2(3))#

#x = (8 +-sqrt(64-60))/6#

#x = (8 +- 2)/6#

For these type of problems, you will obtain two solutions because of the #+-# part. So what you want to do is add 8 and 2 together and divide that by 6:

#x = (8+2)/6#
#x = 10/6 = 1.667#

Now, we subtract 2 from 8 and divide by 6:

#x = (8-2)/6#
# x = 6/6 = 1#

Next, plug each value of x into the equation separately to see if your values give you 0. This will let you know if you performed the calculations correctly or not

Let's try the first value of #x# and see if we obtain 0:

#3(1.667)^(2)-8(1.667)+5 = 0#

#8.33 - 13.33 + 5 =0#

#0= 0#

YAY, this value of x is correct since we got 0!

Now, let's see if the second value of #x# is correct:

#3(1)^(2)-8(1)+5 = 0#

#3 -8 +5 = 0#

#0= 0#
That value of x is correct as well!

Thus, the two possible solutions are:

#x = 1.667 #
#x = 1#