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

Jul 3, 2015

$x = - 4 + i \sqrt{5}$ or $x = - 4 - i \sqrt{5}$

#### Explanation:

Step 1. Write your equation in standard form.

$2 {x}^{2} + 16 x + 42 = 0$

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

Subtract $42$ from each side .

$2 {x}^{2} + 16 x + 42 - 42 = 0 - 42$

$2 {x}^{2} + 16 x = - 42$

Step 3. Divide both sides of the equation by the coefficient of ${x}^{2}$.

Divide both sides by 2.

${x}^{2} + 8 x = - 21$

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

${\left(8\right)}^{2} / 4 = \frac{64}{4} = 16$

Step 5. Add the result to each side.

${x}^{2} + 8 x + 16 = - 21 + 16$

${x}^{2} + 8 x + 16 = - 5$

Step 6. Take the square root of each side.

x+4 = ±isqrt5

Case 1

${x}_{1} + 4 = + i \sqrt{5}$

${x}_{1} = - 4 + i \sqrt{5}$

Case 2

${x}_{2} + 4 = - i \sqrt{5}$

${x}_{2} = - 4 - i \sqrt{5}$

So $x = - 4 + i \sqrt{5}$ or $x = - 4 - i \sqrt{5}$

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

(a) $x = - 4 + i \sqrt{5}$

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) $x = 4 - i \sqrt{5}$

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.