Optional Extra Practice: Section 5.2
Quadratic Word Problems (ZPP)
Solve each word problem by clearly defining a variable, writing an equation, solving the equation,
and thinking carefully about your answer(s). Use ALGEBRA. You will get ZERO credit for “guessing
& checking” on the quiz & test!
1. The square of a number subtracted from 8 times the number is equal to twice the number. Find the
number.
1b) The square of a positive number subtracted from 8 times the number is equal to twice the positive
number. Find the positive number. (Notice this is the same problem as #1, but the word POSITIVE has
been inserted).
2. The product of a “certain negative number decreased by 5” and “the same number tripled and then
increased by 7” is 0. What is the number?
3. Maria said, “The product of my age increased by 4 and 45 less than two times my age is zero. How old
am I?” Find Maria’s age.
4. The product of two consecutive integers is 110. Find the consecutive integers.
5. The length of Charlotte’s garden is 20 yards greater than the width. The area is 300 square yards. How
much fencing would Charlotte need to fence in her garden?
6. The total revenue R earned (in thousands of dollars) from manufacturing calculators is given by the
equation R(p) = -25p2 + 1200p, where p is the price per calculator (in dollars). Find the unit price that
leads to the maximum revenue.
7. One leg of a right triangle is 10 inches less than the length of the hypotenuse. The other leg of the right
triangle is 5 inches less than the hypotenuse. Find the perimeter of this right triangle.
ANSWERS:
1a. The number was 0 or 6.
Equation: 8x – x2 = 2x  ZPP: 0 = x2 – 6x  0 = x(x – 6)  x = 0 or x = 6  (Both #s work – there
were no extraneous solutions)
1b. The number was 6.
Use the same work as #1a above. Our answers were x = 0 or x = 6. BUT 0 is not a positive number, so
it is extraneous. So x = 6 is our final answer.
2. The negative number was -7/3
Equation: (x – 5)(3x + 7) = 0  No need to FOIL, just use ZPP  x = 5 or -7/3  x = -7/3
(x = 5 is extraneous because it’s not negative) 
3. Maria is 22.5 years old.
Equation: (4 + x)(2x – 45) = 0  No need to FOIL, just use ZPP  x = -4 or x = 45/2
 (x = -4 is extraneous)
4. The consecutive integers are “-11 and -10” or “10 and 11”
Equation: x(x + 1) = 110  Distribute and use ZPP: x2 + x – 110 = 0  (x + 11)(x – 10) = 0  x = -11 or
x = 10 (no extraneous solutions here!).
5. The perimeter of her garden is 80 yards.
Use W = width. A = L*W  300 = (20 + w)w  Distribute and use ZPP  0 = w2 + 20w – 300 
0 = (w + 30)(w - 10)  w = 10 (w = -30 is extraneous)  so the length is equal to “20 + w”, so the
length = 20 + (10) = 30 yards. The length is 30 yards and the width is 10 yards. To find the amount of
fencing needed, we need to find the perimeter of her garden. The perimeter would be
2L + 2W = 2(30) + 2(10) = 60 + 20 = 80 yards.
6. The unit price that maximizes their revenue is $24 (Note: unit price means “price per item or unit”…so we
are saying they should sell the calculators for $24 each).
Be careful – this problem is not asking for the x-intercepts! It is asking about the maximum
revenue, which would occur at the vertex of the parabola! Use x = -b/2a to find the x-coordinate of
the vertex, which is what they are asking for here.
7. The perimeter of the triangles is 60 inches.
Solve for the length of the hypotenuse and the legs first! Let x = length of hypotenuse.
Use Pythagorean Theorem: a2 + b2 = c2  (x – 10) 2 + (x – 5) 2 = x2 . Work this out to get:
2x2 – 30x + 125 = x2  x2 – 30x + 125 = 0  (x – 25)(x – 5) = 0  x = 25 or x = 5
 x = 5 is extraneous because it makes a side length negative or zero.  So x = 25 inches
So the sides are 25 inches, 15 inches, and 20 inches.
So the perimeter is 25 + 20 + 15 = 60 inches.