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

2 Answers
Jan 27, 2018

:.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

Jan 27, 2018

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)