Applications of Quadratics: Practice

Consecutive integers are $n$ and $n + 1$, with product $56$.

$n(n + 1) = 56 \;\Rightarrow\; n^2 + n - 56 = 0 \;\Rightarrow\; (n + 8)(n - 7) = 0$

So $n = -8$ or $n = 7$. The integers are positive, so discard $n = -8$; then $n = 7$ and the larger integer is $n + 1 = 8$.

Hitting the water means $h = 0$.

$-16t^2 + 144 = 0 \;\Rightarrow\; t^2 = 9 \;\Rightarrow\; t = \pm 3$

The roots are $t = 3$ and $t = -3$. A time cannot be negative, so discard $t = -3$; the rock hits the water at $t = 3$ seconds.

A width is a length, and a length is never negative.

$w = 6 \ \text{(kept)}, \qquad w = -10 \ \text{(rejected: negative length)}$

So the only physical solution is $w = 6$. The root $w = -10$ satisfies the equation but cannot be a real width.

Let the smaller number be $n$, so the larger is $n + 4$, with product $45$.

$n(n + 4) = 45 \;\Rightarrow\; n^2 + 4n - 45 = 0 \;\Rightarrow\; (n + 9)(n - 5) = 0$

So $n = -9$ or $n = 5$. The numbers are positive, so discard $n = -9$; then the smaller is $5$ and the larger is $9$.

By the Pythagorean theorem, $x^2 + (x + 1)^2 = 5^2$.

$2x^2 + 2x + 1 = 25 \;\Rightarrow\; x^2 + x - 12 = 0 \;\Rightarrow\; (x + 4)(x - 3) = 0$

So $x = -4$ or $x = 3$. A length cannot be negative, so discard $x = -4$; then $x = 3$ (the legs are $3$ and $4$).

Consecutive even integers differ by $2$, so name them $n$ and $n + 2$, with product $48$.

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

So $n = -8$ or $n = 6$. The integers are positive, so discard $n = -8$; the smaller integer is $6$ (and the larger is $8$).

The height model is $h = -16t^2 + v_0 t + h_0$. Here the initial speed is $v_0 = 48$ and the launch height is $h_0 = 0$ (ground level).

$h = -16t^2 + 48t + 0 = -16t^2 + 48t$

Gravity gives the $-16t^2$ term, and the $48$ multiplies $t$ (an initial speed), not a constant.

If one number is $x$, the other is $15 - x$, and their product is $54$.

$x(15 - x) = 54 \;\Rightarrow\; x^2 - 15x + 54 = 0 \;\Rightarrow\; (x - 6)(x - 9) = 0$

So $x = 6$ or $x = 9$, giving the pair $6$ and $9$. Both are positive, and $6 + 9 = 15$ with $6 \times 9 = 54$.

A clock reading during the flight cannot be negative, so a negative root is not a real time.

$t = 3 \ \text{(kept)}, \qquad t = -2 \ \text{(rejected: negative time)}$

The ball lands at $t = 3$ seconds; $t = -2$ would be before the throw and is discarded.

By the Pythagorean theorem, $x^2 + (x + 2)^2 = 10^2$.

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

So $x = -8$ or $x = 6$. A length cannot be negative, so discard $x = -8$; the shorter leg is $6$ (and the longer is $8$).

Consecutive integers are $n$ and $n + 1$, with product $90$.

$n(n + 1) = 90 \;\Rightarrow\; n^2 + n - 90 = 0 \;\Rightarrow\; (n + 10)(n - 9) = 0$

So $n = -10$ or $n = 9$. The integers are positive, so discard $n = -10$; the integers are $9$ and $10$.

Consecutive integers are $n$ and $n + 1$, with product $132$.

$n(n + 1) = 132 \;\Rightarrow\; n^2 + n - 132 = 0 \;\Rightarrow\; (n + 12)(n - 11) = 0$

So $n = -12$ or $n = 11$. The integers are positive, so discard $n = -12$; then $n = 11$ and the larger integer is $12$.