How do you solve the quadratic equation by completing the square: #2x^2 - 7x = 2#?

Question

4231 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-14T16:14:48

The solution is #x = 7/4 +- (sqrt65)/4#.

#2(x^2 - 7/2x) = 2#
#x^2 - 7/2x + 49/16 = 1 + 49/16#
#(x - 7/4)^2 = 65/16#
#(x - 7/4) = +- (sqrt65)/4#

#x = 7/4 +- (sqrt65)/4#

Answer 2 · 2015-07-15T02:03:37

#x=(7+sqrt65)/4,# #(7-sqrt65)/4#

#2x^2-7x=2#

Divide both sides by #2#.

#x^2-7/2x=1#

To complete the square means to force a perfect square trinomial on the left side of the equation in the form #a^2-2ab+b^2=(a-b)^2#.

Divide the coefficient of the #x# term by #2#, square the result, and add to both sides of the equation.

#(-7)/2-:2=(-7/2)*1/2=-7/4#
#(-7/4)^2=49/16#

#x^2-7/2x+49/16=1+49/16#

The common denominator for #1# and #49/16# is #16#. Multiply #1# times #16/16#, then add the two fractions.

#x^2-7/2x+49/16=16/16+49/16# =

#x^2-7/2x+49/16=65/16#

We now have a perfect square trinomial on the left side, where #a=x# and #b=7/4#.

#(x-7/4)^2=65/15#

Take the square root of both sides.

#x-7/4=+-sqrt(65/16)# =

#x-7/4=+-sqrt65/4#

Solve for #x#.

#x=7/4+-sqrt65/4# =

#x=(7+-sqrt65)/4#

#x=(7+sqrt65)/4# =

#x=(7-sqrt65)/4#