Question

How do you solve `(x + 3) ² – (2x – 1)1 = 0`?

Answer 1

`:.x = -2` (plus or minus) `24i`

Explanation:

Rewrite Question:

`(x+3)(x+3) - (2x-1)1 = 0`

Distribute:

`(x^2+3x+3x+9) - (2x-1)`

Collect Like Terms:

`x^2 + 6x + 9 - 2x +1 = 0`

`x^2 +4x + 10 =0`

Use Quadratic Formula:

There are no REAL SOLUTIONS .

You can not square root a NEGATIVE NUMBER.

Therefore:

`:.x = -2` (plus or minus) `24i`

Answer 2

There are two imaginary solutions:

`x=-2+isqrt(6), -2-isqrt(6)`

Explanation:

`(x+3)^2-(2x-1)=0`

`(x+3)^2-2x+1=0`

`x^2+6x+9-2x+1=0`

`x^2+4x+10=0`

Using the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`=>x=(-4+-sqrt((-4)^2-4(1)(10)))/(2(1))`

`=(-4+-sqrt(16-40))/2`

`=(-4+-sqrt(-24))/2`

Since the number under the square root is negative, the solutions are imaginary. You could write "no real solution" for the answer, or if you want the imaginary ones:

`(-4+-sqrt(-24))/2`

`=(-4+-sqrt(-4*6))/2`

`=(-4+-2isqrt(6))/2`

`=-2+-isqrt(6)`

Here are the final imaginary solutions:

`x=-2+isqrt(6), -2-isqrt(6)`