YOU MUST SHOW WORK!
1. A banner is in the shape of a right triangle of area 63in^2 The height of the banner is 4in less than twice the width of the banner. Find the height and width of the banner.
2. A rectangular poster has an area of 190in^2 The height of the poster is 1in less than twice it's width. Find the dimensions of the poster.
1. A banner is in the shape of a right triangle of area 63in^2 The height of the banner is 4in less than twice the width of the banner. Find the height and width of the banner.
2. A rectangular poster has an area of 190in^2 The height of the poster is 1in less than twice it's width. Find the dimensions of the poster.
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Let x = width of the banner. The height, which is "4in less than twice the width" would then be: 2x-4.
Since this is a quadratic equation we want one side to be zero. Subtracting 63 from each side we get:
0= x^2 – 2x - 63
Now we factor (or use the Quadratic Formula). This factors easily:
0 = (x+7)(x-9)
From the Zero Product Property we know that one of these factors must be zero. So:
x+7 = 0 or x-9 = 0
Solving these we get:
x = -7 or x = 9
Since x is the width of the banner and banners do not have negative widths, we will reject the negative solution. So x = 9. And the height, 2x-4, would be 2(9)-4 or 14.
Let x = width of the banner. The height, which is "4in less than twice the width" would then be: 2x-4.
Since this is a quadratic equation we want one side to be zero. Subtracting 63 from each side we get:
0= x^2 – 2x - 63
Now we factor (or use the Quadratic Formula). This factors easily:
0 = (x+7)(x-9)
From the Zero Product Property we know that one of these factors must be zero. So:
x+7 = 0 or x-9 = 0
Solving these we get:
x = -7 or x = 9
Since x is the width of the banner and banners do not have negative widths, we will reject the negative solution. So x = 9. And the height, 2x-4, would be 2(9)-4 or 14.