How do you solve #2x ^ { 2} + 57x = 497#?

2 Answers
Apr 12, 2017

#x = 7# , #x=-35.5#

Explanation:

First we must convert our initial equation so that it fits satisfies # ax^2 + bx + c = 0 #.

#2x^2 + 57x = 497 # becomes #2x^2 + 57x - 497 = 0 #.

From here we conclude the following:

#a=2#
#b=57#
#c=-497#

Now we use the quadratic formula to find the value of #x#.

#x = (-b +- sqrt(b^2 -4ac))/(2a)#

#x = (-57 +- sqrt(57^2 -4(2)(-497)))/(2(2))#

#x = (-57 +- sqrt(3249 -8(-497)))/(4)#

#x = (-57 +- sqrt(3249 + 3976))/(4)#

#x = (-57 +- sqrt(7225))/(4)#

#x = (-57 +- 85)/(4)#

#x = (-57 + 85)/4 # , #x=(-57-85)/4#

#x = (28)/(4) # , # x = (-142)/(4) #

#x = 7# , #x=-35.5#

In this case it is not specified whether we want the positive # x # value, so we are left with two solutions which are #x = 7# and #x=-35.5#.

Apr 12, 2017

Given: #2x^2+57X=497#

We can quickly turn this into an easily recognizable quadratic equation by subtracting 497 from both sides to get:

#2x^2+57X-497=0#

There are a couple of "give aways" we can see on this equation that will take us to the solution:

1) The #x# exponent is #^2# so there are two answers for #x# and there will need to be two brackets in the solution.

2) the multiplier #497# looks like it should be divided by #7#.

#497/7=71#

Setting up the brackets: #(2x ...71)(x ...7)=0#

There is a #+ and -# in the given equation so we know there will be a #+ and -# in the brackets. We also need the larger value to be #+#, so that will be the #71x#:

#(2x + 71)(x - 7)=0# and we see that #71x - 7xx2x = 57x#

Then: #(2x + 71)=0#

#2x = - 71#

#x = - 71/2#

And: #(x-7) = 0#

#x = 7#

To check, put the answers back into the #given# equation:

#2x^2+57x=497#

#2(7)^2+57(7)=497#

#2(49)+399=497#

#98+399=497#

And the other factor:

#2x^2+57x=497#

#2(-71/2)^2+57(-71/2)=497#

#(5041/2) + (-4047/2) = 497#

#5041 - 4047 = 2xx497#

#994=994#