How do you solve by completing the square: #x^2 + 8x + 2 = 0#?

2 Answers
Apr 2, 2015

The answer is #x = -4##+-##sqrt (14)#

The general form of a trinomial is #ax^2+bx+c=0"# The letter #c# is the constant.

Solve the trinomial #x^2+8x+2=0#

First move the constant to the right side by subtracting 2 from both sides.

#x^2+8x=-2#

Divide only the coefficient of 8x by 2. Square the result, and add that value to both sides of the equation.

#((8)/(2))^2=(4)^2=16#

#x^2+8x+16 = -2+16#

#x^2+8x+16=14#

The left side is now a perfect square trinomial. Factor the perfect square trinomial.

#(x+4)^2=14#

Take the square root of each side and solve.

#x+4#=#+-##sqrt (14)#

#x = -4##+-##sqrt (14)#

Source:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

Apr 2, 2015

The answer is #x = -4##+-##sqrt (14)#

The general form of a trinomial is #ax^2+bx+c=0"# The letter #c# is the constant.

Solve the trinomial #x^2+8x+2=0#

First move the constant to the right side by subtracting 2 from both sides.

#x^2+8x=-2#

Divide only the coefficient of 8x by 2. Square the result, and add that value to both sides of the equation.

#((8)/(2))^2=(4)^2=16#

#x^2+8x+16 = -2+16#

#x^2+8x+16=14#

The left side is now a perfect square trinomial. Factor the perfect square trinomial.

#(x+4)^2=14#

Take the square root of each side and solve.

#x+4#=#+-##sqrt (14)#

#x = -4##+-##sqrt (14)#

Source:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm