SOLUTION: I need some help with this problem. I have tried to work it and cannot seem to come up with the answer All the directions say is to complete the square to solve the following equa

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Question 148962: I need some help with this problem. I have tried to work it and cannot seem to come up with the answer All the directions say is to complete the square to solve the following equations
x^2+6x=7 ( i know that you need to get every thing on 1 side but then what)
x^2+6x-7=0

and with this problem the same directions but would you us the quadratic formula to solve? 2x^2-5x-3=0 Thanks Leann
Found 4 solutions by ankor@dixie-net.com, mangopeeler07, MathTherapy, josgarithmetic:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
I need some help with this problem. I have tried to work it and cannot seem to come up with the answer All the directions say is to complete the square to solve the following equations
x^2+6x=7 ( i know that you need to get every thing on 1 side but then what)
x^2+6x-7=0
:
If they want you to solve by completing the square you can leave 7 on the right
And find the value that will complete the square
:
x^2 + 6x + __ = 7
:
To find the term that will complete the square,
divide the coefficient of x by 2 and square that. 6/2 = 3, 3^2 = 9, so we have:
;
x^2 + 6x + 9 = 7 + 9; we have to add 9 to both sides to preserve equality
:
x^2 + 6x + 9 = 16; we know that left side is (x+3)(x+3) or (x+3)^2, so we have:
:
(x + 3)^2 = 16
:
Find the square root of both sides;
x + 3 = +/-sqrt%2816%29
x + 3 = +/-4
Two solutions:
x = -3 + 4
x = +1
and
x = -3 - 4
x = -7
;
You can confirm that: FOIL (x-1)(x+7) = x^2 - 1x + 7x - 7 = x^2 + 6x - 7
:
:
and with this problem the same directions but would you us the quadratic formula to solve? 2x^2 - 5x - 3 = 0
:
you could use the quadratic formula, but this will factor to: (2x + 1)(x - 3)
:
If you insist on using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In your equation: a=2; b=-5; c=-3
x+=+%28-%28-5%29+%2B-+sqrt%28+-5%5E2-4%2A2%2A-3+%29%29%2F%282%2A2%29+
x+=+%285+%2B-+sqrt%2825+-+%28-24%29+%29%29%2F%284%29+
x+=+%285+%2B-+sqrt%2825+%2B+24+%29%29%2F%284%29+
x+=+%285+%2B-+sqrt%2849%29%29%2F%284%29+
Two solutions
x = %285+%2B+7%29%2F4
x = +3
and
x = %285+-+7%29%2F4
x = -1%2F2
:
:
Did this help you? Any questions?

Answer by mangopeeler07(462) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+6x-7=0
Factor:
start with
(x)(x)
insert a + and a -
(x+)(x-)

Find two factors of 7 whose difference is 6: 7 and 1. insert them:
(x+7)(x-1)

Set each expression equal to zero
x+7=0
x-1=0
Solve for x in each
x=-7
x=1

Answer by MathTherapy(10839) About Me  (Show Source):
You can put this solution on YOUR website!
I need some help with this problem.  I have tried to work it and cannot seem to come up with the answer All the directions say is to
complete the square to solve the following equations

x^2+6x=7  ( i know that you need to get every thing on 1 side but then what)
x^2+6x-7=0
and with this problem the same directions but would you us the quadratic formula to solve?   2x^2-5x-3=0  Thanks Leann
*************************************
While one of the other 2 "respondents" solved by factoring the trinomial, this is NOT the required method. However, this method, if the 
trinomial is factorable, is oftentimes used by this author, as a way to determine the solutions/roots before COMPLETING the SQUARE.  

You started out with x%5E2+%2B+6x+=+7, and went on to SUBTRACT 7 from each side of the equation. But, when completing the square, the
CONSTANT (the number without an attached variable) should be on the right side of the equation, which it is! So, NO NEED to subtract 7,
in this case!

The answer to your question about whether or not the quadratic equation formula can be used, is YES!! But, why would you want to do
that when requested to solve, by COMPLETING the SQUARE? And also, both equations are indeed, factorable.
                         x%5E2+%2B+6x+=+7
  ---- Squaring 1%2F2 of b, then adding result to both sides
             x%5E2+%2B+6x+%2B+%28%22%2B+3%22%29%5E2+=+7+%2B+%28%22%2B+3%22%29%5E2 
                         %28x+%2B+3%29%5E2+=+7+%2B+9 
                         %28x+%2B+3%29%5E2+=+16               
                    sqrt%28%28x+%2B+3%29%5E2%29+=+0+%2B-+sqrt%2816%29 ---- Taking square root on both sides
                             x+%2B+3+=+0+%2B-+4 
                                  x+=+-+3+%2B-+4                            
                          

Now, follow the same concept for the 2nd equation!

Answer by josgarithmetic(39823) About Me  (Show Source):
You can put this solution on YOUR website!
______________________
2x%5E2-5x-3=0
______________________

Good choice is Quadratic Formula solution or you could try to factor.

Directly into general formula solution x=%285%2B-+sqrt%285%5E2-4%2A2%2A%28-3%29%29%29%2F%282%2A2%29
%285%2B-+sqrt%2825%2B24%29%29%2F4
%285%2B-+sqrt%2849%29%29%2F4
%285%2B-+7%29%2F4
3 or -1/2


The quadratic trinomial IS factorable but you can do that on your own.

If you wanted to solve by Completing the Square, you can but more effort.
x%5E2-%285%2F2%29x-3%2F2=0
and the term to complete the square is %285%2F%282%2A2%29%29%5E2 which you would add to both sides.