Applications of Quadratics: Additional Practice (Set 2)

Let the height on the wall be $h$; the base distance is $h + 2$, and the ladder is the hypotenuse $10$. By the Pythagorean theorem, $h^2 + (h + 2)^2 = 10^2$.

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

So $h = -8$ or $h = 6$. A height cannot be negative, so the ladder reaches $6$ ft up the wall (the base is $8$ ft out).

Consecutive integers are $n$ and $n + 1$, with product $156$.

$n(n + 1) = 156 \;\Rightarrow\; n^2 + n - 156 = 0 \;\Rightarrow\; (n + 13)(n - 12) = 0$

So $n = -13$ or $n = 12$. The integers are positive, so discard $n = -13$; the smaller integer is $12$ (and the larger is $13$).

Set revenue equal to $1000$ and divide by $-20$.

$p(300 - 20p) = 1000 \;\Rightarrow\; p^2 - 15p + 50 = 0 \;\Rightarrow\; (p - 5)(p - 10) = 0$

So $p = 5$ or $p = 10$; both are positive and keep attendance positive, so both are valid. The lower price is $5$ dollars.

Each outer dimension grows by $2x$, so $(8 + 2x)(10 + 2x) = 143$.

$4x^2 + 36x + 80 = 143 \;\Rightarrow\; 4x^2 + 36x - 63 = 0 \;\Rightarrow\; x = \frac{-36 \pm 48}{8}$

So $x = 1.5$ or $x = -10.5$. A width cannot be negative, so discard $-10.5$; the frame is $1.5$ in wide (the outer rectangle is $11$ by $13$).

Let the integers be $n$ and $n + 1$, so $n^2 + (n + 1)^2 = 313$.

$2n^2 + 2n + 1 = 313 \;\Rightarrow\; n^2 + n - 156 = 0 \;\Rightarrow\; (n + 13)(n - 12) = 0$

So $n = -13$ or $n = 12$. We need positive integers, so discard $n = -13$; then $n = 12$ and the larger is $13$.

Let the height be $h$; the base distance is $h + 3$, and the ladder is the hypotenuse $15$. Then $h^2 + (h + 3)^2 = 15^2$.

$2h^2 + 6h + 9 = 225 \;\Rightarrow\; h^2 + 3h - 108 = 0 \;\Rightarrow\; (h + 12)(h - 9) = 0$

So $h = -12$ or $h = 9$. A height cannot be negative, so the ladder reaches $9$ ft up the wall (the base is $12$ ft out).

Each outer dimension grows by $2x$, so $(9 + 2x)(12 + 2x) = 180$.

$4x^2 + 42x + 108 = 180 \;\Rightarrow\; 2x^2 + 21x - 36 = 0 \;\Rightarrow\; (2x - 3)(x + 12) = 0$

So $x = \tfrac{3}{2}$ or $x = -12$. A width cannot be negative, so the mat is $1.5$ in wide (the outer rectangle is $12$ by $15$).

Half the perimeter is $17$, so length plus width is $17$; write the width as $w$ and the length as $17 - w$. The area is $60$.

$w(17 - w) = 60 \;\Rightarrow\; w^2 - 17w + 60 = 0 \;\Rightarrow\; (w - 5)(w - 12) = 0$

So $w = 5$ or $w = 12$, giving the pair $5$ m by $12$ m. These satisfy both the perimeter ($34$) and the area ($60$).

The area of a right triangle is half the product of its legs. Let the shorter leg be $x$; the longer is $x + 3$.

$\frac{1}{2}x(x + 3) = 14 \;\Rightarrow\; x^2 + 3x - 28 = 0 \;\Rightarrow\; (x + 7)(x - 4) = 0$

So $x = -7$ or $x = 4$. A length cannot be negative, so the shorter leg is $4$ cm (and the longer is $7$ cm).

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

$x = 9 \ \text{(kept)}, \qquad x = -16 \ \text{(rejected: negative length)}$

So the leg is $x = 9$. The reason is that a length cannot be negative, not anything about being odd or even.

Revenue is price times quantity: $p$ times $(200 - 8p)$, set equal to the target of $1200$ dollars.

$p(200 - 8p) = 1200$

Setting it to $0$ asks when revenue vanishes, and $200p - 8p$ drops the required factor of $p$ on the second term.

Consecutive even integers differ by $2$, so name them $n$ and $n + 2$, with product $120$.

$n(n + 2) = 120 \;\Rightarrow\; n^2 + 2n - 120 = 0 \;\Rightarrow\; (n + 12)(n - 10) = 0$

So $n = -12$ or $n = 10$. We need positive integers, so discard $n = -12$; then $n = 10$ and the larger is $12$.