Exercise 44 Page 160

Let x be the side length of a square. If we increase the side length and width by 4 inches, the area of the new square will be (x+4)^2. cc Original Area & New Area x^2 & (x+4)^2 Since the new area is 4 times the original area, we have the equation below. 4x^2=(x+4)^2Let's simplify it and write all the terms on the left hand-side of the equation.

4x^2=(x+4)^2

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

4x^2=x^2+8x+16

SubEqn

LHS-4x^2=RHS-4x^2

0=- 3x^2+8x+16

RearrangeEqn

Rearrange equation

- 3x^2+8x+16=0

Now, we will solve it by using the Quadratic Formula. We first need to identify the values of a, b, and c. - 3x^2+8x+16=0 ⇔ - 3x^2+ 8x+ 16=0 We see that a= - 3, b= 8, and c= 16. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 8±sqrt(( 8)^2-4( - 3)( 16))/2( - 3)

▼

Evaluate right-hand side

CalcPow

Calculate power

x=- 8±sqrt(64-4(- 3)(16))/2(- 3)

MultPosNeg

a(- b)=- a * b

x=- 8±sqrt(64-4(- 3)(16))/- 6

MultNegPos

(- a)b = - ab

x=- 8±sqrt(64-4(- 48))/- 6

MultNegNegOnePar

- a(- b)=a* b

x=- 8±sqrt(64+192)/- 6

AddTerms

Add terms

x=- 8±sqrt(256)/- 6

CalcRoot

Calculate root

x=- 8 ± 16/- 6

The solutions for this equation are x= - 8 ± 16- 6. Let's separate them into the positive and negative cases.

x=- 8 ± 16/- 6
x_1=- 8 + 16/- 6	x_2=- 8 - 16/- 6
x_1=- 8/6	x_2=24/6
x_1 ≈- 1.3	x_2=4

Using the Quadratic Formula, we found that the solutions of the given equation are x_1≈ - 1.3 and x_2=4. Since the length cannot be negative, the side length of the square is 4 inches.