solutionAIAlgebra 2 solves Solve the quadratic

The quadratic formula, applied

Adam Zettel · September 22, 2026 · written with an agent

The ball thrown up at 4 m/s from 1 m lands after 1 s. The quadratic formula gives that, but for $a < 0$ its minus sign gives the larger root, so the solution writes the roots as $h \pm w$ with $w \geq 0$ .

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A ball thrown straight up at 4 m/s from 1 m above the ground has height $y (t) = 1 + 4 t - 5 t^{2}$ . It lands when $- 5 t^{2} + 4 t + 1 = 0$ .

The formula

The roots of $a x^{2} + b x + c = 0$ are

x = \frac{- b \pm b ^{2} - 4 a c}{2 a} .

For the ball, $a = - 5$ , $b = 4$ , $c = 1$ , so $b^{2} - 4 a c = 16 + 20 = 36$ and

t = \frac{- 4 \pm 6}{- 10}, t = - 0.2 or t = 1.

The ball lands after 1 s. The other root, $- 0.2$ s, is a time before the throw.

Which root is smaller

With the minus sign the formula gave $t = 1$ , the larger root: dividing by $2 a = - 10$ flipped the order. Split the formula into a center and a half-width instead,

x = h \pm w,

where $h = - b /2 a$ and $w = b^{2} - 4 a c / 2∣ a ∣$ .

Since $w \geq 0$ , $h - w$ is the smaller root for every $a$ . For the ball, $h = 0.4$ and $w = 0.6$ .

Printed

t1: -0.2 t2: 1

Tests 2 of 2 pass

$x^2 - 5x + 6 = 0$ gives 2 and 3 (passes)
$-5t^2 + 4t + 1 = 0$ gives -0.2 and 1, smaller first (passes)