+101 a) The area of a rectangle is x^2+7x+13. If the length is (x+4), write an expression for the width of the rectangle, including the remainder.
b) If the perimeter of the rectangle is 7, what are the length and width?
+9481 a)
area = length * width
x2 + 7x + 13 = (x + 4) * width Divide both sides of the equation by (x + 4) .
(x2 + 7x + 13)/(x + 4) = width We can do this using long division....
So we get that
width = \(x+3+\frac1{x+4}\)
b)
perimeter = 2[ length + width ]
7 = 2[ x + 4 + x + 3 + \(\frac1{x+4}\) ] Divide both sides by 2 .
3.5 = 2x + 7 + \(\frac1{x+4}\) Subtract 7 from both sides.
-3.5 = 2x + \(\frac1{x+4}\) Multiply through by ( x + 4 )
-3.5(x + 4) = 2x(x + 4) + 1
-3.5x - 14 = 2x2 + 8x + 1 Add 3.5x and 14 to both sides.
0 = 2x2 + 11.5x + 15 We can use the quadratic formula to solve for x.
\(x = {-11.5 \pm \sqrt{132.25-120} \over 4}={-11.5 \pm 3.5 \over 4}=-2.785\pm0.875\)
So, we get that x = -2 and x = -3.75
So......if x = -2
length = -2 + 4 = 2
width = -2 + 3 + 1/(-2+4) = 1 + 1/2 = 1.5
And....if x = -3.75
length = -3.75 + 4 = 0.25
width = -3.75 + 3 + 1/(-3.75+4) = -0.75 + 4 = 3.25
+9481 a)
area = length * width
x2 + 7x + 13 = (x + 4) * width Divide both sides of the equation by (x + 4) .
(x2 + 7x + 13)/(x + 4) = width We can do this using long division....
So we get that
width = \(x+3+\frac1{x+4}\)
b)
perimeter = 2[ length + width ]
7 = 2[ x + 4 + x + 3 + \(\frac1{x+4}\) ] Divide both sides by 2 .
3.5 = 2x + 7 + \(\frac1{x+4}\) Subtract 7 from both sides.
-3.5 = 2x + \(\frac1{x+4}\) Multiply through by ( x + 4 )
-3.5(x + 4) = 2x(x + 4) + 1
-3.5x - 14 = 2x2 + 8x + 1 Add 3.5x and 14 to both sides.
0 = 2x2 + 11.5x + 15 We can use the quadratic formula to solve for x.
\(x = {-11.5 \pm \sqrt{132.25-120} \over 4}={-11.5 \pm 3.5 \over 4}=-2.785\pm0.875\)
So, we get that x = -2 and x = -3.75
So......if x = -2
length = -2 + 4 = 2
width = -2 + 3 + 1/(-2+4) = 1 + 1/2 = 1.5
And....if x = -3.75
length = -3.75 + 4 = 0.25
width = -3.75 + 3 + 1/(-3.75+4) = -0.75 + 4 = 3.25
