you want to build a playhouse for your sister in your backyard. You have the blue print which show the playhouse is 13 feet long and 12 feet wide. you want to change the dimension such that the lenght will be decreased and the width will be increased by the same value. a) write a polynomial that represents the new area of the play room. b) use the discriminant to show that it is possible to change the playroom and have new area of 150 ft^2. c) find the new dimensions of the playroom when its are is 150ft^2.
Hmm..
lets start with the polynomial
(x+13)(-x+12)
= -x^2 - x + 156
part b)
The discriminant is everything inside the square root of the quadratic formula
150= -x^2 - x + 156
-x^2-x+6
a=1
b=-1
c=6
so:
b^2-4*a*c is the discriminant
-1^2 - 4*-1* 6 = 25
and 25 > 0
part c)
Just solve the roots for
150= -x^2 - x + 156
x=-2,3
where 2 is the real solution
a) To represent the new area of the playroom after changing its dimensions, we can start by defining the original length and width. Let's call the original length "l" and the original width "w." According to the blueprint, l = 13 feet and w = 12 feet.
Since we want to change the dimensions such that the length is decreased by the same value the width is increased by, let's use "x" to represent this value. Therefore, the new length would be (l - x) and the new width would be (w + x).
To find the new area, we multiply the new length by the new width. Hence, the polynomial that represents the new area of the playroom is:
New area = (l - x) * (w + x)
Substituting the given values, the polynomial becomes:
New area = (13 - x) * (12 + x)
b) Using the discriminant, we can determine whether it is possible to change the playroom and have a new area of 150 ft^2. The discriminant is calculated using the formula:
Discriminant = b^2 - 4ac
In this case, the quadratic equation representing the new area is:
(New area) = (13 - x)(12 + x)
To simplify, let's rewrite it as a quadratic equation:
(New area) = x^2 - x - 156
Comparing this equation to the general quadratic equation form, ax^2 + bx + c = 0, we can identify:
a = 1
b = -1
c = -156
Now, we can substitute these values into the discriminant formula:
Discriminant = (-1)^2 - 4(1)(-156)
Discriminant = 1 + 624
Discriminant = 625
Since the discriminant (625) is a positive value, it indicates that there are two distinct real solutions for the equation. Therefore, it is indeed possible to change the playroom and have a new area of 150 ft^2.
c) To find the new dimensions of the playroom when its area is 150 ft^2, we can set the new area expression equal to 150 and solve for x:
(New area) = 150
x^2 - x - 156 = 150
Rearranging the equation to standard quadratic form, we have:
x^2 - x - 306 = 0
Now, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
For this equation, a = 1, b = -1, and c = -306. Plugging in these values, we get:
x = (1 ± √((-1)^2 - 4(1)(-306))) / (2*1)
Simplifying further:
x = (1 ± √(1 + 1224)) / 2
x = (1 ± √1225) / 2
x = (1 ± 35) / 2
This gives us two possible values for x:
x₁ = (1 + 35) / 2 = 36 / 2 = 18
x₂ = (1 - 35) / 2 = -34 / 2 = -17
Since dimensions cannot be negative, we discard the second value (-17). Therefore, the new value for x is 18.
Now, we can find the new dimensions of the playroom:
New length = (l - x) = (13 - 18) = -5 feet (discarded)
New width = (w + x) = (12 + 18) = 30 feet
Hence, the new dimensions of the playroom, when its area is 150 ft^2, would be -5 feet (discarded due to not possible) for the length and 30 feet for the width.