How do you solve #2x^2-9x+4=(2x-1)^2#?

1 Answer
Sep 11, 2015

#x = -3# or #x = 1/2#

Explanation:

#2x^2 - 9x + 4 = (2x - 1)^2#

We will start by expanding (FOILing ) the right-hand side of the equation:

#2x^2 - 9x + 4 = 4x^2 - 4x + 1#

Next, subtract #2x^2# from both sides:

#-9x + 4 = 2x^2 - 4x + 1#

Add #9x# to both sides:

#4 = 2x^2 + 5x + 1#

And now, subtract 4 from both sides:

#0 = 2x^2 + 5x - 3#

This is a simple quadratic equation in #x# so we can solve it by completing the square or applying the quadratic formula or whatever method you prefer. I'll apply the quadratic formula:

#x = (-b ± sqrt(b^2 - 4ac))/(2a) = (-5 ± sqrt(5^2 - 4*2*(-3)))/(2*2)#

Simplification gives us:

#x = (-5 ± sqrt(49))/4#

Which further reduces to:

#x = (-5 ± 7)/4#

From here it should be easy to see that

#x = -3# or #x = 1/2#