Area of a Rectangle

A rectangular garden has a length that is 3 meters more than its width. If the area is 70 square meters, find the dimensions.

Define the variable: Let $x$ = width of the garden (in meters) = Length = $x + 3$

Write the area equation: Area = length $\times$ width\n$70 = (x + 3) \times x$ = $70 = x^2 + 3x$

Rearrange to standard form: $x^2 + 3x - 70 = 0$ = Standard form achieved

Factor the quadratic: Find factors of $-70$ that add to $3$:\n$10 \times (-7) = -70$ and $10 + (-7) = 3$ = $(x + 10)(x - 7) = 0$

Apply zero product property: $x + 10 = 0$ or $x - 7 = 0$\n$x = -10$ or $x = 7$ = $x = -10$ or $x = 7$

Check which solution makes sense: Width cannot be negative, so $x = 7$ = Width = 7 m, Length = 10 m

Answer: The garden is 7 meters wide and 10 meters long.

Product of Consecutive Integers

The product of two consecutive positive integers is 156. Find the integers.

Define the variable: Let $x$ = first integer = Second integer = $x + 1$

Write the equation: Product = $x(x + 1) = 156$ = $x^2 + x = 156$

Rearrange to standard form: $x^2 + x - 156 = 0$ = Standard form achieved

Factor the quadratic: Find factors of $-156$ that add to $1$:\n$13 \times (-12) = -156$ and $13 + (-12) = 1$ = $(x + 13)(x - 12) = 0$

Solve for x: $x = -13$ or $x = 12$ = Since integers are positive, $x = 12$

Find both integers: First = 12, Second = $12 + 1 = 13$ = Check: $12 \times 13 = 156$ ✓

Answer: The consecutive integers are 12 and 13.

Projectile Motion

A ball is thrown upward from ground level with an initial velocity of 48 feet per second. Its height $h$ (in feet) after $t$ seconds is given by $h = 48t - 16t^2$. When will the ball hit the ground?

Set up the equation: The ball hits the ground when $h = 0$\n$0 = 48t - 16t^2$ = $48t - 16t^2 = 0$

Factor out the GCF: $16t(3 - t) = 0$ = GCF is $16t$

Apply zero product property: $16t = 0$ or $3 - t = 0$\n$t = 0$ or $t = 3$ = Two solutions found

Interpret the solutions: $t = 0$: when the ball was thrown\n$t = 3$: when it lands = Ball lands at $t = 3$

Answer: The ball will hit the ground after 3 seconds.

Picture Frame Border

A picture that is 8 inches by 12 inches is placed in a frame with a uniform border. If the total area (picture plus frame) is 192 square inches, find the width of the border.

Define the variable: Let $x$ = width of the border (in inches) = Border width is $x$

Find total dimensions: Total width = $8 + 2x$ (border on both sides)\nTotal length = $12 + 2x$ = Dimensions include borders

Write the area equation: $(8 + 2x)(12 + 2x) = 192$ = Total area = 192

Expand and simplify: $96 + 16x + 24x + 4x^2 = 192$\n$4x^2 + 40x + 96 = 192$\n$4x^2 + 40x - 96 = 0$ = Expanded form

Divide by 4 to simplify: $x^2 + 10x - 24 = 0$ = Simplified equation

Factor the quadratic: Find factors of $-24$ that add to $10$:\n$12 \times (-2) = -24$ and $12 + (-2) = 10$ = $(x + 12)(x - 2) = 0$

Apply zero product property: $x + 12 = 0$ or $x - 2 = 0$\n$x = -12$ or $x = 2$ = $x = -12$ or $x = 2$

Check which solution makes sense: Width cannot be negative, so $x = 2$ = Border width = 2 inches

Answer: The border is 2 inches wide.

Mistake: Forgetting to check if the solution makes sense in context

Why: Mathematically, $x = -10$ might be a valid solution to the equation, but a width cannot be negative in real life.

Correct: Always ask: Does this answer make sense? Reject negative values for lengths, times, or quantities that must be positive.

Mistake: Setting up the equation incorrectly

Why: Misinterpreting phrases like '3 more than' or '5 less than' leads to wrong equations.

Correct: '3 more than $x$' means $x + 3$. '5 less than $x$' means $x - 5$. Read carefully and double-check your translation.

Mistake: Forgetting to rearrange to standard form before factoring

Why: You can only use factoring effectively when the equation equals zero.

Correct: Always move all terms to one side: $ax^2 + bx + c = 0$ before attempting to factor.

Mistake: Missing the second border in area problems

Why: A frame border adds to BOTH sides, so total width = original + $2x$, not $x$.

Correct: For borders or margins, remember they're added to both sides. Draw a diagram to visualize.

Architecture and Design

Architects often know the area they need but must calculate dimensions based on constraints.

A room must have 200 square feet of floor space. If the length must be 5 feet more than the width, what are the dimensions?

Sports and Physics

The height of thrown or launched objects follows a quadratic path. Factoring helps find key moments.

A basketball's height (in feet) is $h = -16t^2 + 20t + 6$. When is the ball at 6 feet?

Business and Economics

Profit equations are often quadratic. Finding break-even points uses factoring.

A company's profit is modeled by $P = -x^2 + 50x - 400$, where $x$ is units sold. Find the break-even points (where $P = 0$).

Q: What if the quadratic doesn't factor nicely?

A: Not all word problems result in equations that factor with integers. If you can't find integer factors, you may need to use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Q: How do I know which solution to use when I get two answers?

A: Consider the context. For physical quantities like length, width, time, or count, negative solutions are usually rejected. Choose the solution that makes sense in the real-world situation.

Q: Why do area problems give quadratic equations?

A: Area = length $\times$ width. When both dimensions depend on the same variable $x$ (like $x$ and $x + 3$), multiplying them creates an $x^2$ term, making it quadratic.