Question

Select the correct answer from each drop-down menu.

Answer 1

Answer:

Part 1) The quadratic equation is $x^{2}-5x-50=0$

Part 2) The length of rectangle is 10 in and the width is 5 in

Step-by-step explanation:

Part 1)

Find the quadratic equation

Let

x -----> the length of rectangle

y ----> the width of rectangle

we know that

The area of rectangle is equal to

$A=xy$

$A=50\ in^{2}$

so

$50=xy$ -----> equation A

$x=y+5$

$y=x-5$ -----> equation B

substitute equation B in equation A

$50=x(x-5)\\50=x^{2} -5x\\ x^{2}-5x-50=0$

Part 2) Find the length of the rectangle

$x^{2}-5x-50=0$

Solve the quadratic equation by graphing

The solution is $x=10\ in$

see the attached figure

Find the value of y

$y=10-5=5\ in$

therefore

The length of rectangle is 10 in and the width is 5 in

Answer 2

Answer:

Quadratic equation: $x^{2} +5x-50=0$

Length of the rectangle: 10 inches.

Step-by-step explanation:

In order to solve this you just have to factorize the equation to solve the different values for X:

$x^{2} +5x-50=0\\(x+10)(x-5)=0$

So the only possible answer for the problem would be 5, so if the width is equal to X and the length is x+5 then the length of the rectangle would be 5+5 and that would be 10.