Question

What are the solutions of 3x2 + 15 = −6x

Answer 1

Answer:

x = -1 + 2 i or x = -1 - 2 i

Step-by-step explanation:

Solve for x:

3 (x^2 + 5) = -6 x

Expand out terms of the left hand side:

3 x^2 + 15 = -6 x

Add 6 x to both sides:

3 x^2 + 6 x + 15 = 0

Divide both sides by 3:

x^2 + 2 x + 5 = 0

Subtract 5 from both sides:

x^2 + 2 x = -5

Add 1 to both sides:

x^2 + 2 x + 1 = -4

Write the left hand side as a square:

(x + 1)^2 = -4

Take the square root of both sides:

x + 1 = 2 i or x + 1 = -2 i

Subtract 1 from both sides:

x = -1 + 2 i or x + 1 = -2 i

Subtract 1 from both sides:

Answer: x = -1 + 2 i or x = -1 - 2 i

Answer 2

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTION}}\\\\\boxed{x=-1-2i\ \vee\ x=-1+2i}$

Step-by-step explanation:

$3x^2+15=-6x\qquad\text{add}\ 6x\ \text{to both sides}\\\\3x^2+6x+15=0\qquad\text{divide both sides by 3}\\\\\dfrac{3x^2}{3}+\dfrac{6x}{3}+\dfrac{15}{3}=0\\\\x^2+2x+5=0\qquad\text{subtract 5 from both sides}\\\\x^2+2x=-5\qquad\text{add}\ 1^2\ \text{to both sides}\\\\x^2+2x+1^2=-5+1^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+1)^2=-5+1\\\\(x+1)^2=-4$

$\text{because the square of any real number is not negative.}$

$\text{In the set of complex niumbers:}\\\\i=\sqrt{-1}\\\\(x+1)^2=-4\to x+1=\pm\sqrt{-4}\\\\x+1=\pm\sqrt{(-1)(4)}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x+1=\pm\sqrt{-1}\cdot\sqrt4\\\\x+1=\pm2i\qquad\text{subtract 1 from both sides}\\\\x=-1\pm2i$