SOLUTION: Regarding algebfra, polynomials -- Please help me solve this problem. I know the answer is x = 5, but I'm having trouble arriving at the answer correctly. "The area of a rec

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Regarding algebfra, polynomials -- Please help me solve this problem. I know the answer is x = 5, but I'm having trouble arriving at the answer correctly. "The area of a rec      Log On


   

Question 730838: Regarding algebfra, polynomials --
Please help me solve this problem. I know the answer is x = 5, but I'm having trouble arriving at the answer correctly.
"The area of a rectangle is 55 square feet. The height is x and the length is 2x + 1. Solve for x."
+x%282x+%2B+1%29+=+55+
2x^2 + x = 55

2x^2 + x -55 = 0

My email address is mjj1107@sbcglobal.net.
Thanks.
Mary Jane
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You are on the right path and just need to solve 2x%5E2+%2B+x+-55+=+0+
There are easier quadratic equations

I know of 3 ways to solve quadratic equations like 2x%5E2+%2B+x+-55+=+0+
Factoring will work if the answers are rational numbers.
Completing the square will always work.
Using the quadratic formula will always work too.

FACTORING:
When the leading coefficient is not 1 or -1, factoring is a little harder.
In this case, you must look for pairs of factors of 2%2A55=110 .
Giving a negative sign to one factor and a positive sign to the other, they must add up to the coefficient of the term in x, 1.
The number 110 can be written as 4 different products:
110=1%2A110
110=2%2A55
110=5%2A22
110=10%2A11
The last one is the one that works because 11-10=1
So we use 11 and -10 as coefficients of x and write 2x%5E2+%2B+x+-55 as
2x%5E2+%2B+11x+-10x+-55
Then we factor by grouping, like this:

So, since 2x%5E2+%2B+x+-55=%282x%2B11%29%28x-5%29 we re-write the equation as
%282x%2B11%29%28x-5%29=0 and find the solutions that make
2x%2B11=0 --> x=-11%2F2 and
x-5=q --> x=5
Since x must be positive to be the width of a rectangle, the only solution is x=5%29.

COMPLETING THE SQUARE:
2x%5E2+%2B+x+=+55+ --> x%5E2%2Bx%2F2=55%2F2 dividing both sides by 2
x%5E2%2Bx%2F2 is part of x%5E2%2Bx%2F2%2B1%2F16=%28x%2B1%2F4%29%5E2 so if we add 1%2F16 to both sides we "complete the square:
x%5E2%2Bx%2F2=55%2F2 --> x%5E2%2Bx%2F2%2B1%2F16=55%2F2%2B1%2F16 --> %28x%2B1%2F4%29%5E2=440%2F16%2B1%2F16 --> %28x%2B1%2F4%29%5E2=441%2F16
441%2F16=21%5E2%2F4%5E2=%2821%2F4%29%5E2 or sqrt%28441%2F16%29=sqrt%28441%29%2Fsqrt%2816%29=21%2F4 so
either x%2B1%2F4=21%2F4 --> x=21%2F4-1%2F4 --> x=20%2F4 --> x=5
or x%2B1%2F4=-21%2F4 --> x=-21%2F4-1%2F4 --> x=-22%2F4 --> x=-11%2F2
Same solutions to the equation, and the only solution to the geometry problem is x=5.

THE QUADRATIC FORMULA
is a formula that derives from completing the square.
I never set to memorize it, but I have been using it for so long that I remember it.
For an equation of the form
ax%5E2%2Bbx%2Bc=0 the solutions are given by the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In the case of 2x%5E2+%2B+x+-55+=+0+ a=2 , b=1 and c=-55 so
x+=+%28-1+%2B-+sqrt%281%5E2-4%2A2%2A%28-55%29+%29%29%2F%282%2A2%29+
x+=+%28-1+%2B-+sqrt%281%2B440%29%29%2F4+
x+=+%28-1+%2B-+sqrt%28441%29%29%2F4+
x+=+%28-1+%2B-+21%29%2F4+
So the solutions of the equation are
x=%28-1+%2B+21%29%2F4+ --> x=20%2F4 --> x=5%7D%7D+and%0D%0A%7B%7B%7Bx=%28-1+-+21%29%2F4+ --> x=-22%2F4 --> x=-11%2F2
Since _11%2F2 cannot be the with of a rectangle, the only solution is x=5 .