How do you complete the square to solve #2x^2 + 16x + 42 = 0#?

1 Answer
Jul 3, 2015

Answer:

#x = -4+isqrt5# or #x = -4-isqrt5#

Explanation:

Step 1. Write your equation in standard form.

#2x^2 + 16x +42 = 0#

Step 2. Move the constant to the right hand side of the equation.

Subtract #42# from each side .

#2x^2+16x +42 -42 = 0-42#

#2x^2+16x = -42#

Step 3. Divide both sides of the equation by the coefficient of #x^2#.

Divide both sides by 2.

#x^2 +8x = -21#

Step 4. Square the coefficient of x and divide by 4.

#(8)^2/4 = 64/4 = 16#

Step 5. Add the result to each side.

#x^2 +8x + 16 =-21 +16 #

#x^2 +8x + 16= -5#

Step 6. Take the square root of each side.

#x+4 = ±isqrt5#

Case 1

#x_1 + 4 = +isqrt5#

#x_1 = -4+isqrt5#

Case 2

#x_2 + 4 = -isqrt5#

#x_2 = -4 -isqrt5 #

So #x = -4+isqrt5# or #x = -4-isqrt5#

Check: Substitute the values of #x# back into the quadratic.

(a) #x = -4+isqrt5#

#2x^2 + 16x +42 = 2(-4+isqrt5)^2 + 16(-4+isqrt5) +42 = 2(16 -8isqrt5-5) -64 +16isqrt5 +42 = 32 –cancel(16isqrt5) -10 -64 + cancel(16isqrt5) +42= 0#.

(b) #x = 4 - isqrt5#

#2x^2 + 16x +42 = 2(-4-isqrt5)^2 + 16(-4-isqrt5) +42 = 2(16 +8isqrt5-5) -64 -16isqrt5 +42 = 32 –cancel(16isqrt5) -10 -64 - cancel(16isqrt5) +42= 0#.