Applications of Quadratics: Practice

Set the height to $96$, move it across, and divide by $-16$.

$-16t^2 + 64t - 16 = 0 \;\Rightarrow\; t^2 - 4t + 1 = 0 \;\Rightarrow\; t = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$

Both roots are positive ($\approx 0.27$ and $\approx 3.73$), so both are real times: the ball is $96$ ft high on the way up and again on the way down.

Let the width be $w$; the length is $w + 4$. Then $w(w + 4) = 20$, so $w^2 + 4w - 20 = 0$.

$w = \frac{-4 \pm \sqrt{16 + 80}}{2} = \frac{-4 \pm 4\sqrt{6}}{2} = -2 \pm 2\sqrt{6}$

Since $-2 - 2\sqrt{6} < 0$, that root is a negative length and is discarded. The width is $-2 + 2\sqrt{6}$ m (about $2.9$ m).

Set the height to $20$, giving $-16t^2 + 48t - 20 = 0$, or $4t^2 - 12t + 5 = 0$.

$(2t - 1)(2t - 5) = 0 \;\Rightarrow\; t = 0.5 \ \text{ or }\ t = 2.5$

Both times are valid (going up at $0.5$ s, coming down at $2.5$ s). The question asks when it FIRST reaches $20$ ft, which is the smaller time, $0.5$ s.

If one number is $x$, the other is $10 - x$, and their product is $18$, so $x(10 - x) = 18$, giving $x^2 - 10x + 18 = 0$.

$x = \frac{10 \pm \sqrt{100 - 72}}{2} = \frac{10 \pm \sqrt{28}}{2} = 5 \pm \sqrt{7}$

Both $5 + \sqrt{7}$ and $5 - \sqrt{7}$ are positive, and they are the two numbers (each is the other's partner).

Let the shorter leg be $x$; the longer is $x + 2$, and the hypotenuse is $6$. Then $x^2 + (x + 2)^2 = 36$, so $2x^2 + 4x - 32 = 0$, or $x^2 + 2x - 16 = 0$.

$x = \frac{-2 \pm \sqrt{4 + 64}}{2} = \frac{-2 \pm 2\sqrt{17}}{2} = -1 \pm \sqrt{17}$

Since $-1 - \sqrt{17} < 0$ is not a length, discard it; the shorter leg is $-1 + \sqrt{17}$ cm (about $3.1$ cm).

The equation is $n^2 - 12 = n$, or $n^2 - n - 12 = 0$, and both $4$ and $-3$ satisfy it (indeed $(-3)^2 - 12 = -3$). The deciding factor is the wording "positive number."

$n = 4 \ \text{(kept: positive)}, \qquad n = -3 \ \text{(rejected: not positive)}$

So $n = 4$. The reason is the positivity requirement, not that "negatives cannot be squared" (they can).

Set $h = 0$: $-16t^2 + 20t + 5 = 0$, or $16t^2 - 20t - 5 = 0$. Apply the quadratic formula.

$t = \frac{20 \pm \sqrt{400 + 320}}{32} = \frac{20 \pm 12\sqrt{5}}{32} = \frac{5 \pm 3\sqrt{5}}{8}$

The root $\tfrac{5 - 3\sqrt{5}}{8}$ is negative (since $3\sqrt{5} > 5$), so it is rejected as a negative time. The ball lands at $\tfrac{5 + 3\sqrt{5}}{8}$ s (about $1.5$ s).

Consecutive integers are $n$ and $n + 1$, with product $210$.

$n(n + 1) = 210 \;\Rightarrow\; n^2 + n - 210 = 0 \;\Rightarrow\; (n + 15)(n - 14) = 0$

So $n = -15$ or $n = 14$. Discard the negative root; the integers are $14$ and $15$ (and $14 \times 15 = 210$).

Half the perimeter is $13$, so length plus width is $13$; write the width as $w$ and the length as $13 - w$. The area is $40$.

$w(13 - w) = 40 \;\Rightarrow\; w^2 - 13w + 40 = 0 \;\Rightarrow\; (w - 5)(w - 8) = 0$

So $w = 5$ or $w = 8$, giving the pair $5$ cm by $8$ cm. Both are positive, and they satisfy both the perimeter ($26$) and the area ($40$).

The outer rectangle is $(5 + 2x)(11 + 2x)$, and the path is the outer area minus the pond's $55$ m$^2$.

$(5 + 2x)(11 + 2x) - 55 = 80 \;\Rightarrow\; 4x^2 + 32x - 80 = 0 \;\Rightarrow\; x^2 + 8x - 20 = 0$

Then $(x + 10)(x - 2) = 0$, so $x = -10$ or $x = 2$. A width cannot be negative, so the path is $2$ m wide.

Expand and divide by $-8$.

$200p - 8p^2 = 1200 \;\Rightarrow\; p^2 - 25p + 150 = 0 \;\Rightarrow\; (p - 10)(p - 15) = 0$

So $p = 10$ or $p = 15$; both are positive and keep the quantity positive, so both are valid. The higher price is $15$ dollars.

Let the even integers be $n$ and $n + 2$, so $n^2 + (n + 2)^2 = 100$.

$2n^2 + 4n + 4 = 100 \;\Rightarrow\; n^2 + 2n - 48 = 0 \;\Rightarrow\; (n + 8)(n - 6) = 0$

So $n = -8$ or $n = 6$. We need positive integers, so discard $n = -8$; then $n = 6$ and the larger is $8$ (and $36 + 64 = 100$).