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:
[tex]\large\boxed{\bold{NO\ REAL\ SOLUTION}}\\\\\boxed{x=-1-2i\ \vee\ x=-1+2i}[/tex]
Step-by-step explanation:
[tex]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<0\qquad\bold{NO\ REAL\ SOLUTION}[/tex]
[tex]\text{because the square of any real number is not negative.}[/tex]
[tex]\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[/tex]
