How do you solve #2x ^ { 2} + 57x = 497#?
2 Answers
Explanation:
First we must convert our initial equation so that it fits satisfies
#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 = (-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
Given:
We can quickly turn this into an easily recognizable quadratic equation by subtracting 497 from both sides to get:
There are a couple of "give aways" we can see on this equation that will take us to the solution:
1) The
2) the multiplier
Setting up the brackets:
There is a
Then:
And:
To check, put the answers back into the
And the other factor: