SOLUTION: the outside of a picture frame has a length which is 5cm more than the width, the area enclosed of the picture frame is 126cm what is the width of the outside

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: the outside of a picture frame has a length which is 5cm more than the width, the area enclosed of the picture frame is 126cm what is the width of the outside      Log On


   

Question 1190852: the outside of a picture frame has a length which is 5cm more than the width, the area enclosed of the picture frame is 126cm what is the width of the outside
Found 3 solutions by ikleyn, Alan3354, greenestamps:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
the outside of a picture frame has a length which is 5cm more than the width,
the area enclosed of the picture frame is 126cm
what is the width of the outside 

    w*(w+5) = 126


    w^2 + 5w - 126 = 0


    w%5B1%2C2%5D = %28-5+%2B-+sqrt%285%5E2+-4%2A1%2A%28-126%29%29%29%2F2 = %28-5+%2B-+sqrt%28529%29%29%2F2 = %28-5%2B-+23%29%2F2.


We discard the negative root and accept the positive one.


ANSWER.  The width is  %28-5+%2B+23%29%2F2 = 9 cm.

Solved.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
the outside of a picture frame has a length which is 5cm more than the width, the area enclosed of the picture frame is 126cm what is the width of the outside
--------------------
Find a pair of factors of 126 that differ by 5.
It's easy since they're integers.

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


An attempt to solve the problem using formal algebra leads to this:

x%28x%2B5%29=126
x%5E2%2B5x=126
x%5E2%2B5x-126=0

To solve that by factoring, you need to find two integers whose product is 126 and whose difference is 5. But that is what the original problem asks you to do, so the formal algebra doesn't get you any closer to the answer.

So if formal algebra is required and you aren't going to finish the solution by finding two integers whose product is 126 and whose difference is 5, then you have to use the quadratic formula; and that is just plugging numbers into a formula, which doesn't teach you anything or give you any mental exercise.

So do what the other tutor says: just find the two integers whose product is 126 and whose difference is 5. Do that by looking at all the pairs of integers with a product of 126:

126 = 126 * 1
126 = 63 * 2
126 = 42 * 3
126 = 21 * 6
126 = 14 * 9 <-- there it is

ANSWER: width 9 (and length 14)