SOLUTION: The length of a rectangle is 3 ft more than twice the width, and the area of the rectangle is 77 ft2 . Find the dimensions of the rectangle. ---once i start to work out th

Question 987508: The length of a rectangle is 3 ft
more than twice the width, and the area of the rectangle is 77 ft2
. Find the dimensions of the rectangle.

---once i start to work out the problem i get stuck at 2w^2+3w-77=0. I know 7*-11= -77 but do you not use the ac method to make it add up to 3?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
L length
w width
Description:
, the area.

You have the right equation. Either solve by factoring or by direct use of the formula for solution of a quadratic equation. Maybe you need just to look at all the combinations for decomposing into factors!

(2x 7)(x 11) or (2x 11)(x 7)

Which will give you what you want?
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 3 ft
more than twice the width, and the area of the rectangle is 77 ft2
. Find the dimensions of the rectangle.

---once i start to work out the problem i get stuck at 2w^2+3w-77=0. I know 7*-11= -77 but do you not use the ac method to make it add up to 3?

Yes, you're indeed correct. You should use the "ac" method to factor.
The "ac" method "tells" you that you need 2 factors that multiply to - 154 (2 * - 77), and sum to + 3 (b)
These factors are: + 14 and - 11
You then get:
You then factor out the GCF in each of the 2 expressions: "" and ""
This will give you the binomial factors of the trinomial, which you set equal to 0,
and subsequently, you will obtain the roots/zeroes/solutions of the quadratic equation.
Obviously if a negative value is calculated for w, you will ignore that negative value, as measurements must be > 0.
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