How do you solve #2x^2=x+28 #?

3 Answers
Aug 3, 2015

Answer:

Move all values and variables to the left and either factor, solve with the quadratic equation, or solve by completing the square; you could also leave the expression alone and solve by graphing.

Explanation:

#2x^2=x+28#
#2x^2-x=28#
#2x^2-x-28=0#

Next, try to factor this expression.

#(2x+a)(x+b)=0#

Where #a xx b=-28# and #2bx + ax=-x# or #2b+a=-1#

The factors of -28 are:
#-(a xx b)# or #-(b xx a)#
#-(1 xx 28)#
#-(2 xx 14)#
#-(4 xx 7)#

Of these factors, only the last pair will allow you to make #2b+a=-1# by setting #b=-4# and #a=7#. But this means we can factor the expression above.

#(2x+a)(x+b)=0#
#(2x+7)(x-4)=0# which is only true when

#x=-7/2# or #x=4#.

Check your work by plugging these values back into the original expression.

Aug 3, 2015

Answer:

#x_(1,2) = 1/4 +- 15/4#

Explanation:

You could solve this quadratic equation by completing the square.

To do that, start by getting your quadratic to the general form

#color(blue)(x^2 + b/ax = -c/a)#

by adding #-x# to both sides of the equation and dividing your terms by #2#

#2x^2 - x = color(red)(cancel(color(black)(x))) - color(red)(cancel(color(black)(x))) + 28#

#(color(red)(cancel(color(black)(2))) * x^2)/(color(red)(cancel(color(black)(2)))) - x/2 = 28/2#

#x^2 - x/2 = 14#

Now, you need to add a term to both sides of the equation so that the left side of the equation can be written as the square of a binomial.

The coefficient of the #x#-term will tell what term you need to add. More specifically, divide this coefficient by #2# and square the result to get

#(-1/2 * 1/2)^2 = 1/16#

Add #1/16# to both sides of the equation

#x^2 - x/2 + 1/16 = 14 + 1/16#

The left side of the equation can now be rewritten as

#x^2 - 2 * (1/4) * x + (1/4)^2 = (x-1/4)^2#

You now have

#(x-1/4)^2 = 225/16#

Take the square root of both sides of the equation

#sqrt( (x-1/4)^2) = sqrt(225/16)#

#x-1/4 = +- 15/4 => x_(1,2) = 1/4 +- 15/4#

The two solutions to the equation will thus be

#x_1 = 1/4 + 15/4 = color(green)(4)# and #x_2 = 1/4 - 15/4 = color(green)(-7/2)#

Aug 4, 2015

Answer:

Solve #y = 2x^2 - x - 28 = 0# (1)

Explanation:

I use the new Transforming Method.
Transformed equation #y' = x^2 - x - 56 = 0 #(2).
Factor pairs of (-56) --> (-4, 14)(-7, 8). This sum is 1 = -b. Two real roots of (2) are: -7 and 8.
The 2 real roots of equation (1) are: #(-7/2)# and #(8/2 = 4)#