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

2 Answers
Joseph
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#

Aviv S.
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)#

Related questions
See all questions in Comparing Methods for Solving Quadratics
Impact of this question
14897 views around the world
You can reuse this answer
Creative Commons License