Question

The length of a rectangle is 10 inches more than 2 times the width. If the area of the rectangle is 28 square

Answer 1

Answer:≥2

Step-by-step explanation:

First, we need to translate the problem into algebraic expressions and equations:

[length][is][ten inches more than][twice the width]

⟹[][=][10+][2×]

⟹=10+2

==(10+2)()=22+10

[Area][is at least][28]

[][≥][28]

⟹≥28

Putting these together, we have a quadratic inequality:

22+10≥28

The most straightforward way [I know of] to solve this is to compare to zero, find the roots, and go from there:

22+10−28≥0

⟹2+5−14≥0

Finding the zeros using your favorite method — factoring, quadratic formula, etc. — yields zeros of  ={−7,2} .

Since the width of the rectangle can’t be negative, our solution involves  =2 . The graph of the quadratic function is concave up (due to the positive leading coefficient), meaning the function is non-negative for  ≤−7  or  ≥2  — giving us our solution,  ≥2 !