Applications of Quadratic Equations

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a \ne 0\). Many real situations naturally produce this shape: an object thrown into the air, a rectangular garden with a fixed perimeter, a price change that affects both quantity sold and revenue. In each case, one quantity is related to the square of another, so a quadratic equation is exactly the right tool to describe it and solve for the unknown value.

The challenge in these problems is rarely the algebra itself. It is translating a sentence into an equation. Once you have the equation, you can use the same techniques you already know, such as factoring, the quadratic formula, or completing the square.

1. Read carefully and identify the unknown. Assign it a variable, such as \(x\) or \(t\).
2. Translate the relationships into an equation. Look for phrases like "product of," "area of," "sum of the squares," or a given formula such as height versus time.
3. Write the equation in standard form, \(ax^2 + bx + c = 0\), so it is ready to solve.
4. Solve the equation using factoring, completing the square, or the quadratic formula.
5. Check both solutions against the context. A quadratic often has two roots, but only one (or neither) may make sense as an answer to the original question.

A ball is launched upward from a platform. Its height in feet after \(t\) seconds is given by \(h(t) = -16t^2 + 64t + 80\). When does the ball hit the ground?

The ball hits the ground when \(h(t) = 0\), so we solve \(-16t^2 + 64t + 80 = 0\). Dividing every term by \(-16\) gives \(t^2 - 4t - 5 = 0\), which factors as \((t - 5)(t + 1) = 0\). This gives \(t = 5\) or \(t = -1\). Since time cannot be negative, the ball hits the ground at \(t = 5\) seconds.

The graph below shows the height of the ball over time, including its starting height, its highest point, and the moment it lands.

A gardener has 40 meters of fencing to enclose a rectangular plot against a wall, so fencing is only needed on three sides. If the two equal sides have length \(x\), the side parallel to the wall is \(40 - 2x\), and the area is \(A = x(40 - 2x)\). If the gardener wants an area of 200 square meters, we solve:

\(x(40 - 2x) = 200\)

\(40x - 2x^2 = 200\)

\(-2x^2 + 40x - 200 = 0\)

Dividing by \(-2\): \(x^2 - 20x + 100 = 0\), which factors as \((x - 10)^2 = 0\), so \(x = 10\) meters. This is the one size that gives exactly 200 square meters here, since the discriminant is zero. You can confirm this using the discriminant to see why only one repeated solution exists.

Find two consecutive positive integers whose product is 132. Let the smaller integer be \(x\), so the next one is \(x + 1\). Then \(x(x + 1) = 132\), which expands to \(x^2 + x - 132 = 0\). Factoring gives \((x - 11)(x + 12) = 0\), so \(x = 11\) or \(x = -12\). Since the integers must be positive, \(x = 11\), and the integers are 11 and 12.

A shop sells a product for \(\$20\), moving 300 units per week. For every \(\$1\) price increase, weekly sales drop by 10 units. If \(x\) is the number of \(\$1\) increases, revenue is \(R(x) = (20 + x)(300 - 10x)\). Setting \(R(x) = 6000\) gives:

\(6000 + 100x - 10x^2 = 6000\)

\(-10x^2 + 100x = 0\)

\(-10x(x - 10) = 0\)

So \(x = 0\) or \(x = 10\). This means the current price already gives \(\$6000\) revenue, and raising the price by \(\$10\) gives the same revenue again, since the revenue curve rises and then falls back down.

Once your equation is set up, choose whichever method fits the numbers best. Simple, factorable equations are quickest by factoring, while messier coefficients are usually easier with the quadratic formula. Whatever method you choose, always finish by checking each solution against the real-world limits of the problem, such as time, length, or price being non-negative.

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