Applications of Quadratics: Practice

The frame adds $x$ on each side, so each outer dimension grows by $2x$. Set the outer area to $63$.

$(5 + 2x)(7 + 2x) = 63 \;\Rightarrow\; 4x^2 + 24x + 35 = 63$

Move $63$ across and divide by $4$: $x^2 + 6x - 7 = 0$, so $(x + 7)(x - 1) = 0$ and $x = -7$ or $x = 1$. A width cannot be negative, so the frame is $1$ in wide.

Set the height to $48$ and divide by $-16$.

$-16t^2 + 64t = 48 \;\Rightarrow\; t^2 - 4t + 3 = 0 \;\Rightarrow\; (t - 1)(t - 3) = 0$

So $t = 1$ or $t = 3$. Both are positive, so both count: the ball is $48$ ft high at $1$ s (going up) and again at $3$ s (coming down).

Let the integers be $n$ and $n + 1$, so $n^2 + (n + 1)^2 = 85$.

$2n^2 + 2n + 1 = 85 \;\Rightarrow\; n^2 + n - 42 = 0 \;\Rightarrow\; (n + 7)(n - 6) = 0$

So $n = -7$ or $n = 6$. We need positive integers, so discard $n = -7$; the integers are $6$ and $7$ (and $36 + 49 = 85$).

Set revenue equal to $350$ and divide by $-10$.

$p(120 - 10p) = 350 \;\Rightarrow\; p^2 - 12p + 35 = 0 \;\Rightarrow\; (p - 5)(p - 7) = 0$

So $p = 5$ or $p = 7$; both are positive and keep the quantity positive, so both are valid. The lower price is $5$ dollars.

By the Pythagorean theorem, $x^2 + (x + 7)^2 = 13^2$.

$2x^2 + 14x + 49 = 169 \;\Rightarrow\; x^2 + 7x - 60 = 0 \;\Rightarrow\; (x + 12)(x - 5) = 0$

So $x = -12$ or $x = 5$. A length cannot be negative, so discard $x = -12$; the shorter leg is $5$ (and the longer is $12$).

During a flight the clock cannot read a negative value, so a negative root is not a real time.

$t = 5 \ \text{(kept)}, \qquad t = -1.5 \ \text{(rejected: negative time)}$

The physical answer is $t = 5$ seconds; the root $t = -1.5$ solves the equation but is discarded.

Let the width be $w$; the length is $w + 2$. Set the area to $48$.

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

So $w = -8$ or $w = 6$. Discard the negative width; the width is $6$ m, and the question asks for the length, which is $w + 2 = 8$ m.

By the Pythagorean theorem, $x^2 + (x + 14)^2 = 26^2$.

$2x^2 + 28x + 196 = 676 \;\Rightarrow\; x^2 + 14x - 240 = 0 \;\Rightarrow\; (x + 24)(x - 10) = 0$

So $x = -24$ or $x = 10$. Discard the negative leg; the shorter leg is $10$, and the longer leg is $x + 14 = 24$.

Set revenue equal to $480$ and divide by $-5$.

$p(100 - 5p) = 480 \;\Rightarrow\; p^2 - 20p + 96 = 0 \;\Rightarrow\; (p - 8)(p - 12) = 0$

So $p = 8$ or $p = 12$. The problem gives $8$ dollars, so the other price with the same revenue is $12$ dollars.

Each outer dimension grows by $2x$, so the outer area is $(6 + 2x)(8 + 2x) = 80$.

$4x^2 + 28x + 48 = 80 \;\Rightarrow\; x^2 + 7x - 8 = 0 \;\Rightarrow\; (x + 8)(x - 1) = 0$

So $x = -8$ or $x = 1$. A width cannot be negative, so the border is $1$ cm wide (the outer photo is $8$ by $10$).

"$6$ less than its square" is $n^2 - 6$, and that equals the number $n$.

$n^2 - 6 = n \;\Rightarrow\; n^2 - n - 6 = 0 \;\Rightarrow\; (n - 3)(n + 2) = 0$

So $n = 3$ or $n = -2$. Both solve the equation, but the problem asks for a positive number, so discard $n = -2$; the number is $3$.

Both numbers satisfy the equation, but $w$ stands for a width, and a width cannot be negative.

$w = 6 \ \text{(kept)}, \qquad w = -8 \ \text{(rejected: negative length)}$

So the width is $6$; the root $w = -8$ is a real solution of the equation but not of the situation.