Applications of Quadratics: Additional Practice (Set 1)

Let the width be $w$; the length is $w + 6$. Set the area to $40$.

$w(w + 6) = 40 \;\Rightarrow\; w^2 + 6w - 40 = 0 \;\Rightarrow\; (w + 10)(w - 4) = 0$

So $w = -10$ or $w = 4$. A width cannot be negative, so discard $w = -10$; the width is $4$ m (and the length is $10$ m).

By the Pythagorean theorem, $x^2 + (x + 7)^2 = 17^2$.

$2x^2 + 14x + 49 = 289 \;\Rightarrow\; x^2 + 7x - 120 = 0 \;\Rightarrow\; (x + 15)(x - 8) = 0$

So $x = -15$ or $x = 8$. A length cannot be negative, so discard $x = -15$; the shorter leg is $8$ (and the longer is $15$).

Set the height to $64$ and divide by $-16$.

$-16t^2 + 80t = 64 \;\Rightarrow\; t^2 - 5t + 4 = 0 \;\Rightarrow\; (t - 1)(t - 4) = 0$

So $t = 1$ or $t = 4$. Both are positive, so both count: the ball is $64$ ft high at $1$ s (going up) and again at $4$ s (coming down).

If one number is $x$, the other is $12 - x$, and their product is $35$.

$x(12 - x) = 35 \;\Rightarrow\; x^2 - 12x + 35 = 0 \;\Rightarrow\; (x - 5)(x - 7) = 0$

So $x = 5$ or $x = 7$, giving the pair $5$ and $7$ (and $5 + 7 = 12$ with $5 \times 7 = 35$).

Set revenue equal to $1500$ and divide by $-25$.

$p(400 - 25p) = 1500 \;\Rightarrow\; p^2 - 16p + 60 = 0 \;\Rightarrow\; (p - 6)(p - 10) = 0$

So $p = 6$ or $p = 10$; both are positive and keep the ticket count positive, so both are valid. The lower price is $6$ dollars.

"$20$ less than its square" is $n^2 - 20$, and that equals $n$.

$n^2 - 20 = n \;\Rightarrow\; n^2 - n - 20 = 0 \;\Rightarrow\; (n - 5)(n + 4) = 0$

So $n = 5$ or $n = -4$. Both solve the equation, but the problem asks for a positive number, so discard $n = -4$; the number is $5$.

Set $h = 40$, giving $-16t^2 + 40t - 35 = 0$, or $16t^2 - 40t + 35 = 0$. Check the discriminant.

$D = (-40)^2 - 4(16)(35) = 1600 - 2240 = -640 < 0$

The discriminant is negative, so the roots are complex and there is no real time at which the height is $40$ ft. The drone never reaches that height.

The area of a right triangle is half the product of its legs. Let the shorter leg be $x$; the longer is $x + 7$.

$\frac{1}{2}x(x + 7) = 30 \;\Rightarrow\; x^2 + 7x - 60 = 0 \;\Rightarrow\; (x + 12)(x - 5) = 0$

So $x = -12$ or $x = 5$. Discard the negative length; the shorter leg is $5$ cm, and the question asks for the longer leg, $x + 7 = 12$ cm.

Let the integers be $n$ and $n + 1$, so $n^2 + (n + 1)^2 = 221$.

$2n^2 + 2n + 1 = 221 \;\Rightarrow\; n^2 + n - 110 = 0 \;\Rightarrow\; (n + 11)(n - 10) = 0$

So $n = -11$ or $n = 10$. We need positive integers, so discard $n = -11$; then $n = 10$ and the larger is $11$.

Set $h = 0$: $-16t^2 + 24t + 8 = 0$, or dividing by $-8$, $2t^2 - 3t - 1 = 0$. Apply the quadratic formula.

$t = \frac{3 \pm \sqrt{9 + 8}}{4} = \frac{3 \pm \sqrt{17}}{4}$

The root $\tfrac{3 - \sqrt{17}}{4}$ is negative (since $\sqrt{17} > 3$), so it is rejected as a negative time. The ball lands at $\tfrac{3 + \sqrt{17}}{4}$ s (about $1.8$ s).

Revenue is price times quantity: $p$ times $(150 - 5p)$, set equal to the target of $1000$ dollars.

$p(150 - 5p) = 1000$

Setting it to $0$ asks when revenue vanishes, and $150p - 5p$ drops the required factor of $p$ on the second term.

Expand and divide by $-5$.

$150p - 5p^2 = 1000 \;\Rightarrow\; p^2 - 30p + 200 = 0 \;\Rightarrow\; (p - 10)(p - 20) = 0$

So $p = 10$ or $p = 20$. Both are positive and keep the membership positive (at $10$ dollars, $100$ members; at $20$ dollars, $50$ members), so both prices give a revenue of $1000$ dollars.