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the area of a rectangular cloth is (6x squared-19x-85)cm squared. the length is (2x=5)cm. find the width.

2006-12-14 13:15:26 · 6 answers · asked by nisha10mabry 3 in Science & Mathematics Mathematics

6 answers

To work this out you have (6x^2 - 19x - 85) = (2x - 5)(ax + b)

-5b must equal -85
b = 17

2a must equal 6
a = 3

(6x^2 - 19x - 85) = (2x - 5)(3x + 17)

Check

-15x + 34x = 19x correct

So (2x - 5)(3x + 17) is width times length

width = (3x + 17)cm

2006-12-14 13:26:46 · answer #1 · answered by Tom :: Athier than Thou 6 · 1 0

Remember that the area of a rectangle is the following:

A = Length times Width.

We're given the length is 2x+5, and we're given the area A is 6x^2 - 19x - 85.

Therefore, if we call the width W, and if

A = LW

Then

6x^2 - 19x - 85 = (2x + 5)W

Divide both sides by 2x + 5 to get

[6x^2 - 19x - 85]/[2x + 5] = W

To solve this, we need to factor the numerator. Through trial and error, you'll find that it factors into (2x + 5) (3x - 17). Therefore

{(2x + 5) (3x - 17)} / [2x + 5] = W

Factors on the numerator and denominator cancel each other out, leaving you with

3x - 17 = W, which is our width.

HOWEVER, we must keep in mind that during the steps of our algebra, we divided by (2x + 5) to get where we wanted to go. Division by 0 is a big no-no (and is certainly not allowed in math), so we have give an added restriction: 2x + 5 cannot equal 0. If we solve for x in the equation 2x + 5 = 0, then
2x + 5 = 0
2x = -5
x = -5/2

Therefore, our solution would be W = 3x - 17, for x not equal to -5/2.

Do we need this added restriction? Maybe not, because we know that width is positive. That is,

3x - 17 >= 0, so
3x >= 17, and
x >= 17/3

Our values for x are greater than 17/3 anyway (so there's no need for a restriction).

2006-12-14 13:33:07 · answer #2 · answered by Puggy 7 · 0 0

You wrote 2x = 5 instead of 2x + 5...

6x^2 - 19 x - 85 =

I have to found 2 numbers so that their difference is 19 and their product is 510.

- 85 x 6 = - 510. This - sign means that one of the numbers have to be negative

Considering that the difference, the numbers must be pretty close to each other.

510 = 1 x 510 = 2 x 255 = 3 x 170 = 5 x 102 = 6 x 85 = 10 x 51 = 15 x 34, etc.

510 = 2 x 3 x 5 x 17

15 x 34 seems to be OK. 34 - 15 = 19

So, now do this:

6x^2 - (34 - 15) x - 85 = 6x^2 + 15 x - 34 x - 85 = 3x (2 x + 5) - 15 (2x+5)

Since 2x + 5 is a common factor, then the result is (3x-15)(2x+15)

Anabel

2006-12-14 13:36:01 · answer #3 · answered by Anonymous · 0 0

width=area/length
=6x^2-19x-85/2x+5

2006-12-14 13:25:43 · answer #4 · answered by raj 7 · 0 0

if 2x+5 it is 3x-17. Follow Tom's example above but be careful that you check to multiple the two factors back together

2006-12-14 13:30:11 · answer #5 · answered by samsosa2000 2 · 0 0

Is that supposed to be 2x+5 or 2x-5

2006-12-14 13:28:08 · answer #6 · answered by Film XPress 1 · 0 0

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