Answer:
4. Option C (3a+2) inches
5. Extraneous solution x=-5/6 because we get for width and length negative values.
Solution: Value of x is 3
Length of the box: 5 ft
Width of the box: 4 ft
Step-by-step explanation:
4. Area of a rectangle: A=12a^2-a-6 square inches
Width: w=4a-3
Length: l=?
A=w l
Replacing A by 12a^2-a-6 and w by 4a-3
12a^2-a-6 = (4a-3) l
Solving for l: Dividing both sides of the equation by 4a-3:
(12a^2-a-6) / (4a-3) = (4a-3) l / (4a-3)
Simplifying:
(12a^2-a-6) / (4a-3) = l
l = (12a^2-a-6) / (4a-3)
Factoring the numerator:
12a^2-a-6 = (4a-3)(3a+2)
Let's check it:
(4a-3)(3a+2)=4a(3a)+4a(2)-3(3a)-3(2)=12a^2+8a-9a-6→(4a-3)(3a+2)=12a^2-a-6
Replacing the numerator:
l = (4a-3)(3a+2) / (4a-3)
Simplifying:
l = (3a+2) inches
5. Length: l=(3x-5) ft
Width: w=(2x-1) ft
Height: h=2 ft
Volumen of the box: V=40 ft^3
x=?
Length: l=?
Width: w=?
V = l w h
Replacing the given:
40 ft^3 = (3x-5) ft (2x-1) ft 2 ft
40 ft^3 = 2 (3x-5)(2x-1) ft^3
40=2(3x-5)(2x-1)
Dividing both sides of the equation by 2:
40/2=2(3x-5)(2x-1)/2
Simplifying:
20=(3x-5)(2x-1)
Eliminating the parentheses on the right side of the equation applying the distributive property:
20=3x(2x)+3x(-1)-5(2x)-5(-1)
20=6x^2-3x-10x+5
Adding like terms:
20=6x^2-13x+5
Equaling to zero: Subtracting 20 from both sides of the equation:
20-20=6x^2-13x+5-20
0=6x^2-13x-15
6x^2-13x-15=0
ax^2+bx+c=0; a=6, b=-13, c=-15
Using the quadratic formula:
x=[-b+-sqrt(b^2-4ac)]/(2a)
x=[-(-13)+-sqrt((-13)^2-4(6)(-15))]/(2(6))
x=[13+-sqrt(169+360)]/12
x=[13+-sqrt(529)]/12
x=[13+-23]/12
x1=(13-23)/12=(-10)/12=-10/12=-(10/2)/(12/2)→x1=-5/6
x2=(13+23)/12=36/12→x2=3
With x=-5/6
l=(3x-5) ft
l=(3(-5/6)-5) ft
l=(-5/2-5) ft
l=-(5/2+5) ft
l=-(5+2(5))/2 ft
l=-(5+10)/2 ft
l=-15/2 ft < 0. The length cannot be a negative number then x=-5/6 is a extraneous solution.
w=(2x-1) ft
w=(2(-5/6)-1) ft
w=(-5/3-1) ft
w=-(5/3+1) ft
w=-(5+3(1))/3 ft
w=-(5+3)/3 ft
w=-8/3 ft <0. The width cannot be a negative number then x=-5/6 is a extraneous solution.
With x=3
l=(3x-5) ft
l=(3(3)-5) ft
l=(9-5) ft
l=4 ft
w=(2x-1) ft
w=(2(3)-1) ft
w=(6-1) ft
w=5 ft
and h=2 ft
Let's check the volume
V= w l h
V=(5 ft)(4 ft)(2 ft)
V=40 ft^3 Correct
and 0 doesn't belong to the set.
.