SOLUTION: I was just wondering if someone could tell me if I did this correctly or not and if not how to do it correctly. Thank you. Solve for x without using decimals: (2x + 3)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I was just wondering if someone could tell me if I did this correctly or not and if not how to do it correctly. Thank you. Solve for x without using decimals: (2x + 3)      Log On


   

Question 16324: I was just wondering if someone could tell me if I did this correctly or not and if not how to do it correctly. Thank you.
Solve for x without using decimals:

(2x + 3)^2 - (x - 2)^2 = 3
(2x+3)(2x+3) – (x-2)(x-2) = 3
4x^2 + 6x + 6x + 9 – x^2 – 2x – 2x + 4 = 3
4x^2 + 12x + 9 – x^2 – 4x + 4 = 3
4x^2 + 8x + 13 – x^2 = 3
3x^2 + 8x + 13 = 3
3x^2 + 8x + 10 = 0
A = 3
B = 8
C = 10
8^2 – 4*3*10 = -56
No real solution.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you started out ok, but when you got to the point of removing the second set of parentheses, you apparently forgot that it was preceeded by a negative sign which affects the sign of every term within these parentheses. So when you take the parentheses away, you must change the sign of each term within.
Solve for x:
%282x%2B3%29%5E2+-+%28x-2%29%5E2+=+3 Perform the indicated multiplications.
%284x%5E2+%2B+12x+%2B+9%29+-+%28x%5E2+-4x+%2B+4%29+=+3 Simplify. Remove the parentheses.
4x%5E2+%2B+12x+%2B+9+-+x%5E2+%2B+4x+-+4+=+3 Collect like-terms.
3x%5E2+%2B+16x+%2B+5+=+3 Subtract 3 from both sides.
3x%5E2+%2B+16x+%2B+2+=+0 Use the quadratic formula to solve for x. x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x+=+%28-16%2B-sqrt%2816%5E2+-+4%283%29%282%29%29%29%2F2%283%29
x+=+%28-16%2B-sqrt%28256+-+24%29%29%2F6
x+=+%28-16%2B-sqrt%28232%29%29%2F6
The discriminant is positive (+232) so the solution has two real roots.
The roots are:
x+=+%28-8%2F3%29+%2B+%281%2F3%29sqrt%2858%29
x+=+%28-8%2F3%29+-+%281%2F3%29sqrt%2858%29