Question

Candy draws a square design with a side length of x inches for the window at the pet shop. She takes the design to the printer and asks for a sign that has an area of 16x2 – 40x + 25 square inches. What is the side length, in inches, of the pet shop sign?

Answer 1

Answer:

The length of the sign is $(4x-5)$ inches

Step-by-step explanation:

We are given that,

Area of the square design = $16x^2-40x+25$

We will first find the roots of the equation $16x^2-40x+2=0$

The roots of the quadratic equation $ax^2+bx+c=0$ are given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

On comparing, a= 16, b= -40 and c= 25

So, the roots of the equation are,

$x=\frac{40\pm\sqrt{(-40)^2-4\times 16\times 25}}{2\times 16}$

i.e. $x=\frac{40\pm\sqrt{1600-1600}}{32}$

i.e. $x=\frac{40\pm\sqrt{0}}{32}$

i.e. $x=\frac{40}{32}$

i.e. $x=\frac{5}{4}$

That is, the factors of the polynomial $16x^2-40x+25$ are $(4x-5)$ and $4x-5$.

So, Area of the square design = $16x^2-40x+25$ = $(4x-5)^2$

Since, Area of a square = $Length^2$

Thus, the length of the sign is $(4x-5)$ inches

Answer 2

Answer: The side length of the pet shop sign=5/4 inches

Step-by-step explanation:

Given: Candy draws a square design with a side length of x inches for the window at the pet shop.

She takes the design to the printer and asks for a sign that has an area of $16x^2-40x+25$square inches.

Let y(x) =$16x^2-40x+25$

differentiating both sides w.r.t x we get

y'(x)=32x-40

to find critical point x ,Put y'(x)=0 then

⇒32x-40=0

⇒x=5/4

Therefore the side length of the pet shop sign=5/4 inches.