Applications of Quadratics: Free Response

On the ground means $h = 0$:

$-16t^2 + 32t + 128 = 0.$

All three coefficients share $-16$; dividing by it changes no roots. Term by term, since a negative divisor flips every sign: $-16t^2 \div (-16) = t^2$, $32t \div (-16) = -2t$, $128 \div (-16) = -8$.

$t^2 - 2t - 8 = 0.$

Monic, so find two integers with product $-8$ and sum $-2$: $-4$ and $+2$.

$t^2 - 2t - 8 = (t - 4)(t + 2) = 0 \quad\Rightarrow\quad t = 4 \ \text{ or } \ t = -2.$

Reject $t = -2$: it is two seconds BEFORE the throw, and the model only describes the ball in the air. Check the survivor:

$-16(4)^2 + 32(4) + 128 = -256 + 128 + 128 = 0.$

The ball lands $4$ seconds after it is thrown.

Substituting $t = 0$ gives the height at the throw:

$h = -16(0)^2 + 32(0) + 128 = 128.$

So $128$ feet is the roof. The $32$ multiplies $t$: the feet the ball would rise each second without gravity, its initial upward speed. ($-16t^2$ is gravity, pulling it back.)

$t = -2$ solves the equation, but the equation models the FLIGHT, which begins at $t = 0$. A negative time reads from before the throw, when the model says nothing about the ball, so it cannot be a landing time. Only the meaning of $t$ rules it out; the algebra never could.

The path meets each dimension at BOTH ends: cross the finished rectangle and you cross path, garden, path. A strip of width $x$ at each end adds $2x$ to a side, not $x$, so the outer rectangle measures $(8 + 2x)$ metres by $(12 + 2x)$ metres.

The given $192$ square metres is that outer rectangle's area, the product of its two sides:

$(8 + 2x)(12 + 2x) = 192.$

Expand in full, every term of the first bracket meeting every term of the second:

$(8 + 2x)(12 + 2x) = 96 + 16x + 24x + 4x^2 = 4x^2 + 40x + 96.$

Set equal to $192$ and gather: $4x^2 + 40x - 96 = 0$. All three coefficients share $4$; dividing makes the trinomial monic and factorable:

$x^2 + 10x - 24 = 0 \quad\Rightarrow\quad (x + 12)(x - 2) = 0 \quad\Rightarrow\quad x = -12 \ \text{ or } \ x = 2.$

Discard $x = -12$: a width is a length, never negative. Check the survivor: the outer rectangle is $8 + 4 = 12$ metres by $12 + 4 = 16$ metres, and $12 \times 16 = 192$, as given.

The path is $2$ metres wide.

The whole argument is the SIGN of the constant term. Set the product of the outer dimensions to the total:

$(a + 2x)(b + 2x) = A \quad\Rightarrow\quad 4x^2 + 2(a + b)x + (ab - A) = 0.$

Divide by $4$, since the product-of-roots fact needs a leading coefficient of $1$:

$x^2 + \frac{a + b}{2}\,x + \frac{ab - A}{4} = 0.$

The roots multiply to $\dfrac{ab - A}{4}$. Since $A > ab$, $ab - A < 0$, so that constant term is negative.

The roots are real: complex conjugates multiply to $p^2 + q^2$, never negative. A product of two real numbers is negative exactly when they have opposite signs, so one root is positive and one negative, and since a width must be positive, exactly one root is usable. That proves the first half.

The second half is not fair. The argument settles the COUNT of usable answers before any factoring, but says nothing about their VALUE: it cannot tell you the path is $2$ metres wide. The negative root is not wasted ink either; its negativity is what certifies the other root as the one the situation keeps.

Revenue is price times quantity, so the target gives:

$p(90 - 15p) = 120 \quad\Rightarrow\quad 90p - 15p^2 = 120.$

Gather on the left and divide by $-15$, which also makes the leading coefficient $1$. A negative divisor flips every sign: $-15p^2 \div (-15) = p^2$, $90p \div (-15) = -6p$, $-120 \div (-15) = +8$.

$-15p^2 + 90p - 120 = 0 \quad\Rightarrow\quad p^2 - 6p + 8 = 0.$

Factors with $-2$ and $-4$, product $8$, sum $-6$:

$p^2 - 6p + 8 = (p - 2)(p - 4) = 0 \quad\Rightarrow\quad p = 2 \ \text{ or } \ p = 4.$

Test each root against the story. At $2$ dollars the bakery sells $90 - 15(2) = 60$ muffins; at $4$ dollars, $90 - 15(4) = 30$. Both prices and both quantities are positive, so neither is forbidden, and both hit the target:

$2 \cdot 60 = 120 \qquad \text{and} \qquad 4 \cdot 30 = 120.$

The bakery takes $120$ dollars at $2$ dollars a muffin, selling $60$, and again at $4$ dollars, selling $30$.

A counterexample needs one case where the rule's conclusion fails, and the bakery is one.

At $2$ dollars the bakery sells $90 - 15(2) = 60$ muffins; at $4$ dollars, $90 - 15(4) = 30$. Each is a positive price selling a positive number of muffins, so neither is ruled out, and each hits the target exactly:

$2(90 - 30) = 120 \qquad \text{and} \qquad 4(90 - 60) = 120.$

Both roots survive, so a student obeying the rule would report half the answer.

The rule breaks at 'always'. It confuses a root that is IMPOSSIBLE with one that is merely SECOND: a negative length or time is discarded because no length is negative and no clock reads before the flight began, not because a quadratic comes with a spare root. When the situation forbids nothing, nothing is discarded.

Two prices work here because the cheaper one sells more muffins and the dearer one fewer, multiplying out to the same money.

For a monic quadratic the roots add to the negative of the linear coefficient and multiply to the constant term, so read both off $p^2 - 6p + 8 = 0$:

$p_1 + p_2 = 6 \qquad \text{and} \qquad p_1 p_2 = 8.$

Two numbers with sum $6$ and product $8$ are $2$ and $4$, the pair factoring produced. No coincidence: hunting for $(p - 2)(p - 4)$ IS hunting for two numbers with that sum and product.

The routes differ in what they hand you. Factoring ends with the roots themselves and works whenever they are rational. Sum and product cost one glance at the coefficients and never produce a root; they give information ABOUT the pair, which can settle the interpretation before any solving. The product $8$ is positive, so the roots share a sign; the sum $6$ is positive, so that sign is positive. Neither root will be thrown out as a negative price, and the sum also places the two prices symmetrically about $3$ dollars.

So: factor when you need the numbers, and use sum and product when you need a property of them, above all their signs, before ten minutes of algebra rather than after.

Line 1 is sound: one unknown is named, the other leg written in the same letter, and the Pythagorean theorem applied correctly.

Line 2 is where it breaks: the square of a binomial is not the sum of the squares.

$(x + 2)^2 = (x + 2)(x + 2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4.$

The student wrote $x^2 + 4$, losing the cross term $4x$. Everything after is faithful arithmetic on a wrong equation, so line 2 is the FIRST error and the only one worth repairing:

$x^2 + x^2 + 4x + 4 = 100, \qquad \text{that is} \qquad 2x^2 + 4x + 4 = 100.$

The restored cross term makes this a genuine quadratic; a square root no longer finishes it. Gather onto one side and divide by $2$:

$2x^2 + 4x + 4 = 100 \quad\Rightarrow\quad 2x^2 + 4x - 96 = 0 \quad\Rightarrow\quad x^2 + 2x - 48 = 0.$

Factor with $+8$ and $-6$, product $-48$, sum $+2$:

$(x + 8)(x - 6) = 0 \quad\Rightarrow\quad x = -8 \ \text{ or } \ x = 6.$

Reject $x = -8$: a leg is a length, never negative. Check against the theorem:

$6^2 + 8^2 = 36 + 64 = 100 = 10^2.$

The shorter leg is $x = 6$ centimetres, the longer $x + 2 = 8$ centimetres.

A positive answer is not a correct one. Rejecting a negative root tests MEANING; substituting into the original statement tests TRUTH. The student's answer passes the first and fails the second.

Line 2 quietly replaced the problem: from there the student solves $2x^2 + 4 = 100$, which describes no triangle in the question, and lines 3 to 5 are all faithful to THAT equation. The mistake can never be caught by checking later lines against each other, only by returning to the original sentence.

So substitute into the original Pythagorean relation. If the shorter leg were $x \approx 6.93$, the longer would be $x + 2 \approx 8.93$, and

$6.93^2 + 8.93^2 \approx 48.0 + 79.7 = 127.7,$

while the hypotenuse squared is $10^2 = 100$. These disagree: legs of $6.93$ and $8.93$ demand a hypotenuse of about $\sqrt{127.7} \approx 11.3$ centimetres, not the given $10$. The answer contradicts the problem it claims to solve, condemning it before anyone knows which line went astray.

The correct legs pass the same test exactly: $6^2 + 8^2 = 100 = 10^2$.

Level with the ring means the height equals $30$:

$-16t^2 + 40t + 6 = 30 \quad\Rightarrow\quad -16t^2 + 40t - 24 = 0.$

Every coefficient shares $-8$; dividing leaves numbers small enough to factor:

$2t^2 - 5t + 3 = 0 \quad\Rightarrow\quad (2t - 3)(t - 1) = 0 \quad\Rightarrow\quad t = \frac{3}{2} \ \text{ or } \ t = 1.$

Both times are positive, so neither is forbidden. Check the later one: $-16(1.5)^2 + 40(1.5) + 6 = -36 + 60 + 6 = 30$.

The firework rises past the ring at $t = 1$ second and falls back past it at $t = 1.5$ seconds.

Whether any real time solves the equation is decided by the discriminant alone; the roots are never needed. Set the height to $40$ and gather to standard form, dividing by $-2$ to keep the coefficients whole:

$-16t^2 + 40t - 34 = 0 \quad\Rightarrow\quad 8t^2 - 20t + 17 = 0.$

One line settles it:

$D = (-20)^2 - 4(8)(17) = 400 - 544 = -144.$

Since $D < 0$, $\sqrt{-144} = 12i$ enters the numerator, and the roots are $t = \dfrac{20 \pm 12i}{16} = \dfrac{5}{4} \pm \dfrac{3}{4}i$, complex. A clock cannot read a complex number of seconds, so no real time gives a height of $40$ feet: the firework never reaches the banner.

That is not a failure of the algebra but a report about the world.

Carry the target as a letter and push it through the discriminant. With $a = 16$, $b = -40$, $c = H - 6$:

$D = (-40)^2 - 4(16)(H - 6) = 1600 - 64(H - 6) = 1600 - 64H + 384 = 1984 - 64H.$

A real time exists exactly when $D \ge 0$, LINEAR in $H$:

$1984 - 64H \ge 0 \quad\Longleftrightarrow\quad 64H \le 1984 \quad\Longleftrightarrow\quad H \le 31.$

So no height above $31$ feet is attained at any real time: the greatest height is $31$ feet. At $H = 31$ the discriminant is zero, so the two times collapse into one, $t = \dfrac{40}{32} = 1.25$ seconds, the one height passed only once.

A real root only counts if it is a real TIME, so check these times are not negative. Take a target in the band $6 \le H \le 31$, at or above the platform and not above the ceiling. The upper bound gives $D \ge 0$, so the times are real; the lower bound gives $D = 1984 - 64H \le 1984 - 384 = 1600$, so $\sqrt{D} \le 40$ and both roots

$t = \frac{40 \pm \sqrt{D}}{32}$

are nonnegative. Every target from the platform up to $31$ feet is genuinely reached, at genuine times. The upper bound matters: drop it and the claim is false, since $H = 40$ sits above the platform and is never reached, as part B showed.

The earlier answers agree. The ring at $30$ feet satisfies $30 \le 31$, so it is reached twice, one time either side of $1.25$ seconds. The banner at $40$ feet satisfies $40 > 31$, so it is never reached, exactly what the negative discriminant in part B reported.