SOLUTION: Hello, I’d like some help with this please. Could someone tell me if I have this correct and help me with the last steps of factoring into binomials and using the zero product p

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Question 1031921: Hello,
I’d like some help with this please. Could someone tell me if I have this correct and help me with the last steps of factoring into binomials and using the zero product property?
Thank you!
Phoebe

The length of a rectangular window is 5 feet more than its width. The area of the window is 36 square feet. Find both the length and width of the window.
What equation can be used to find the dimensions of the window?
Put that equation in standard form.
Factor the quadratic into two binomials.
Use the zero product property to solve the equation.
2w+(2w+5)=36
4w^+10w-36=0
Am I on the right track so far?

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Length L and width w;



Simple substitution gives you .
Work with it through all the processes needed for the exercise. Most important is to understand enough to first obtain the equation just shown.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Hello,
I’d like some help with this please. Could someone tell me if I have this correct and help me with the last steps of factoring into binomials and using the zero product property?
Thank you!
Phoebe

The length of a rectangular window is 5 feet more than its width. The area of the window is 36 square feet. Find both the length and width of the window.
What equation can be used to find the dimensions of the window?
Put that equation in standard form.
Factor the quadratic into two binomials.
Use the zero product property to solve the equation.
2w+(2w+5)=36
4w^+10w-36=0
Am I on the right track so far? 
Apparently, you have the width as W

Therefore, length = W + 5

With area being 36 sq ft, we get: W(W + 5) = 36, NOT 2W + (2W + 5) = 36

No, you're NOT on the right track

Quadratic to solve: . 

Solve for W, the width, then add 5 to W to get length 

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