Question

1. a rectangle has a length (x + 4) cm and a width of (x - 2) cm. its area is 112 square centimeters. write a quadratic equation and solve it to find the length and width of the rectangle. show or explain how you got your answer. (2 points)

dimensions:
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(0003.psd.algii.quad.pre.scr-lv2-dok2-cc.9-12.a.ced.1)

1. find the minimum value of the given function. show or explain how you got your answer. (2 points)

f(x) = 2x² + 12x - 18
(0003.psd.algii.quad.pre.scr-lv1-dok1-cc.9-12.f.if.7

1. in simplest form, √(-175) is equivalent to which of the following? show or explain your work. (2 points)

a. 5i√7
b. 7√(-5)
c. 7i√5
d. -5√7
(0008.psd.algii.quad.pre.mc-dok1-cc.9-12.n.c

Explanation:

Step1: Write the quadratic equation

The area of a rectangle is given by \(A = length\times width\). Here, \(length=(x + 4)\) and \(width=(x - 2)\), and \(A = 112\). So, \((x + 4)(x - 2)=112\).
Using the FOIL method: \(x^{2}-2x+4x - 8=112\).
Simplify to get the quadratic equation: \(x^{2}+2x-120 = 0\).

Step2: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (here \(a = 1\), \(b = 2\), \(c=-120\)), we can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(2)^{2}-4\times1\times(-120)=4 + 480=484\).
Then, \(x=\frac{-2\pm\sqrt{484}}{2}=\frac{-2\pm22}{2}\).
We have two solutions:

• \(x=\frac{-2 + 22}{2}=\frac{20}{2}=10\)
• \(x=\frac{-2-22}{2}=\frac{-24}{2}=-12\)

Since \(x\) represents a dimension, \(x>0\), so \(x = 10\).

Step3: Find the length and width

Length \(=x + 4=10+4 = 14\) cm.
Width \(=x - 2=10-2 = 8\) cm.

Answer:

The length of the rectangle is \(14\) cm and the width is \(8\) cm.