Can someone please explain how to get the answer i've looked it up several times and i'm still stuck. I'm in algebra 2 with trig and doing my summer math packet: "The length of a rectangle is two feet less than four times the width. Find the length and width if the area is 38.2 square feet. let w= width of the rectangle" Btw websites say w is 19.1 but my answer key says its 3.35. Thanks!

Step-by-step explanation:

Let L be the length of this rectangle

The length of a rectangle is 2 feet less than four times the width

● L+2 = 4w

The area of this rectangle is 38.2 ft^2

● L*w = 38.2

The system of equations is

L+2 = 4w => L-4w =2

L*w = 38.2 => L= 38.2/w

Replace L by 38.2/w in L-4w =2

● L-4w = 2

● (38.2/w)-4w = 2

●(38,2-4w^2)/w = 2

● 38.2-4w^2 = 2w

● 38.2-4w^2-2w = 0

● -4w^2-2w+38.2 =0

Multiply by -1 to reduce the - signs

● 4w^2+2w-38.2 =0

This is a quadratic equation so we will use the discriminant

□□□□□□□□□□□□□□□□□

The discriminant is b^2-4ac

● b= 2 (2w)

● a= 4 (4w^2)

● c= -38.2 (the constant term)

b^2-4ac =2^2-4*4*(-38.2) = 615.2 > 0

The discriminant is positive so we have two solutions w and w' :

●●●●●●●●●●●●●●●●●●●●●●●●

w= (-2-24.8)/8 = -3.35

24.8 is the root square of 615.2(the discriminant)

●w is negative

● a distance is always positive so this value isn't a solution

w'= (-2+24.8)/8 =2.85 > 0

So this value is a solution for our equation

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w= 2.85 feet

w=3.35

Step-by-step explanation:

The length of a rectangle is two feet less than four times the width:

length=4W-2

A=L*W

A=(4W-2)(W)

38.2=4W^2-2W

4W^2-2W-38.2=0  complete the square to find the solution add term(b/2)²

4W²-2W+1/4=38.2+1/4

4(W²-W/2+1/16)=38.45

4(w-1/4)^2=38.45

(w-1/4)²=38.45/4=9.6125

w-1/4=+ or -√9.6125 ( since it is width it has to be positive

w=√9.6125+1/4 = 3.35