Quadratic Equation Solver guide

How to Solve Quadratic Word Problems

Set up a quadratic word problem by naming your variable, writing an equation from the situation, and rearranging to ax² + bx + c = 0. Solve as usual, then check both roots against the real situation — negative times, lengths or quantities are usually rejected.

The general approach

  • Name the unknown. Write down exactly what x represents, including its units.
  • Build the equation from the relationship the question describes.
  • Rearrange so one side is zero.
  • Solve by factoring or the formula.
  • Check both roots against the situation and discard any that make no physical sense.

Projectile motion

A ball is thrown upward from a 25 metre cliff at 20 m/s. Its height is h = −5t² + 20t + 25, where t is seconds. When does it hit the ground?

Hitting the ground means h = 0, so solve −5t² + 20t + 25 = 0.

Divide everything by −5 first to make life easier: t² − 4t − 5 = 0.

That factors as (t − 5)(t + 1) = 0, giving t = 5 or t = −1.

t = −1 is rejected — the ball cannot land one second before it was thrown. So it lands after 5 seconds.

The equation loaded into the calculator is the original −5t² + 20t + 25, so you can see the same roots come out of the formula directly.

The negative-a clue

In that problem a = −5, which is negative, so the parabola opens downward. That matches the physical situation: the ball goes up, reaches a peak, and comes back down.

The peak is at the vertex: t = −20 ÷ (2 × −5) = 2 seconds. Substituting back gives h = −5(4) + 40 + 25 = 45 metres.

So "what is the maximum height?" is a vertex question, and "when does it land?" is a roots question. Recognising which one is being asked is most of the work.

Area problems

A rectangular garden is 3 metres longer than it is wide, with an area of 40 m². Find its dimensions.

Let the width be x. Then the length is x + 3, and the area is x(x + 3) = 40.

Expanding: x² + 3x = 40, so x² + 3x − 40 = 0.

Factoring: (x + 8)(x − 5) = 0, giving x = −8 or x = 5.

x = −8 is rejected — a width cannot be negative. So the width is 5 m and the length is 8 m.

Check: 5 × 8 = 40. Correct.

Deciding which root to discard

Nearly every quadratic word problem produces two roots and expects you to keep only one. These are the usual grounds for rejecting one:

  • Negative time — before the event started.
  • Negative length, width or distance — physically impossible.
  • Negative quantity of items sold or produced.
  • A value outside a stated range — such as a price above a given maximum.

Say why you rejected it

When you discard a root, write one short sentence explaining why. "Rejected because time cannot be negative" is usually worth a mark on its own.

It also protects you. Occasionally both roots are valid — a ball thrown from ground level passing a given height twice, once on the way up and once coming down. Getting into the habit of justifying the rejection means you notice the cases where there is nothing to reject.

FAQ

Related questions

How do you set up a quadratic word problem?

Name the unknown with its units, write an equation from the relationship described, then rearrange so one side is zero.

Why do I get two answers when the question wants one?

Because a quadratic has two roots, but usually only one makes physical sense. Discard any negative time, length or quantity.

How do I find the maximum height of a projectile?

Find the vertex. Use t = −b ÷ 2a for the time of the peak, then substitute that back to get the height.

Can both roots ever be valid?

Yes. A projectile passing a given height does so twice — once rising and once falling — so both times are real answers.

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