Answer:
no real solutions, [tex]x = \pm2i - 4[/tex]
Step-by-step explanation:
We can solve for x in the equation:
[tex]x(x+8)=-20[/tex]
by completing the square.
First, we can apply the distributive property to the left side:
[tex]x(x) + x(8) = -20[/tex]
[tex]x^2 + 8x = -20[/tex]
Next, we can add (8/2)² to both sides:
[tex]x^2 + 8x + \left(\frac{8}2\right)^2 = -20 + \left(\frac{8}2\right)^2[/tex]
This simplifies as: [tex]4^2 = 16[/tex], so the equation becomes:
[tex]x^2 + 8x + 16 = -4[/tex]
We can now factor the left side as a perfect square trinomial:
[tex]\left(x + \frac{8}{2}\right)^2 = -4[/tex]
[tex](x + 4)^2 = -4[/tex]
Finally, we can take the square root of both sides to solve for x.
[tex]x + 4 = \pm\sqrt{-4}[/tex]
We can see that the solution for x involves the square root of a negative, which is not a real number, so this equation has no real solutions, meaning it doesn't cross the x-axis.
But, we can find an imaginary solution using the imaginary number i:
[tex]i = \sqrt{-1}[/tex]
↓↓↓
[tex]x + 4 = \pm\sqrt{-1}\sqrt4[/tex]
[tex]x + 4 =\pm i\sqrt4[/tex]
[tex]x + 4 = \pm2i[/tex]
[tex]\boxed{x = \pm2i - 4}[/tex]
