View Full Version : Formula issues: expressing length, width, given area formula

Mathaxl

07-10-2006, 10:34 PM

Hi, I hope anyone that is on now can be helpful enough to help me out a bit here. To be more specific, I have been staring at this problem for about 30 minutes trying to figure out what the first step is!

So if someone can be so kind as to tell me what I should do first for this problem?

Thank you in advance.

61. The formula for the area of a rectangle is A = l w. Find possible algebraic expressions for l and w if A = 6x<4> + 5x<2> – 4.

--I just realized that the exponents do not show on here... so I'l put <> around the exponents so y'all can differenciate the numbers apart.

Mrspi

07-11-2006, 12:03 AM

Hi, I hope anyone that is on now can be helpful enough to help me out a bit here. To be more specific, I have been staring at this problem for about 30 minutes trying to figure out what the first step is!

So if someone can be so kind as to tell me what I should do first for this problem?

Thank you in advance.

61. The formula for the area of a rectangle is A = l w. Find possible algebraic expressions for l and w if A = 6x<4> + 5x<2> – 4.

--I just realized that the exponents do not show on here... so I'l put <> around the exponents so y'all can differenciate the numbers apart.

Ok...for this rectangle,

A = 6x<sup>4</sup> + 5x<sup>2</sup> - 4

Can you factor the expression on the right side? If you can write this as the product of two factors, then one factor could be the expression for the length, and the other for the width.

Hint: you may wish to let u = x<sup>2</sup> and write your expression as

6u<sup>2</sup> + 5u - 4

You may find it easier to factor in this form; when you have completed the factoring in terms of u, you can replace u with x<sup>2</sup>.

Denis

07-12-2006, 12:01 AM

A = 6x^4 + 5x^2 - 4

Happens that you can factor the right side directly: (3x^2 + 4)(2x^2 - 1).

So you can have length = 3x^2 + 4 and width = 2x^2 - 1.

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