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

1 Answer
Jul 3, 2015

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.