Quadratic Inequalities: Free Response

Standard form gives the coefficients, factored form gives the zeros, so change the form. Two numbers with product $-28$ and sum $-3$ are $-7$ and $4$.

$x^2 - 3x - 28 = (x - 7)(x + 4)$

The zero product property makes the roots $x = 7$ and $x = -4$, and those two numbers cut the line into the pieces $x < -4$, $-4 < x < 7$ and $x > 7$.

Now take the factors one at a time, which is the whole reason for changing form. The factor $x + 4$ is zero at $-4$, negative for every $x$ below $-4$ and positive for every $x$ above it. The factor $x - 7$ is zero at $7$, negative below $7$ and positive above it.

Put the two readings on the same line. For $x < -4$ both factors are negative. For $-4 < x < 7$ the factor $x + 4$ has already turned positive while $x - 7$ has not, so the two signs disagree. For $x > 7$ both are positive.

A product of two numbers is negative exactly when the two disagree in sign, and positive exactly when they agree. Part A found them disagreeing only on the middle piece.

$(x - 7)(x + 4) < 0 \iff -4 < x < 7$

The symbol is strict, so neither root belongs: at $x = -4$ and at $x = 7$ the product is $0$, which is not less than $0$. The solution set is the open interval $(-4, 7)$.

For $\ge 0$ take instead the two pieces where the factors agree, the ones outside the roots, and this time keep the roots themselves, since a non-strict symbol also accepts $0$.

$x^2 - 3x - 28 \ge 0 \iff x \le -4 \ \text{ or } \ x \ge 7$

That is two separate rays, not one interval, so it is written with a union: $(-\infty, -4] \cup [7, \infty)$. Collapsing it to $[-4, 7]$ would report very nearly the complement, the stretch where the quadratic is negative rather than where it is at or above $0$.

The argument is about the factors, not about the product.

A linear factor $x - r$ is zero at $r$, negative for every $x < r$ and positive for every $x > r$, so it changes sign at $r$ and nowhere else. Both factors of $f$ have that shape, so between two consecutive roots neither one can turn over.

$f(x) = (x - 7)(x + 4)$

Take any piece the roots cut out. It contains no root, so across that whole piece $x - 7$ holds one sign and $x + 4$ holds one sign. A product of two numbers whose signs are fixed has a fixed sign, because $(+)(+)$ and $(-)(-)$ are positive while $(+)(-)$ is negative. So $f$ has one constant sign there, and any single interior value reveals it.

A root is a different matter. There the product is exactly $0$, which is neither positive nor negative, so it reports on neither of the pieces beside it. That is why a test value has to be taken strictly inside a piece: an endpoint describes the boundary, not the interior.

Read the coefficients with their own signs and substitute inside parentheses, which is where a negative coefficient usually gets lost.

For $f$, $a = 3$, $b = -6$ and $c = 7$.

$\Delta = (-6)^2 - 4(3)(7) = 36 - 84 = -48$

A negative discriminant means no real roots, and $a = 3 > 0$ means the parabola opens upward.

For $g$, $a = 16$, $b = -56$ and $c = 49$.

$\Delta = (-56)^2 - 4(16)(49) = 3136 - 3136 = 0$

A zero discriminant means one repeated root, and it sits at the vertex, $x = -\dfrac{b}{2a} = \dfrac{56}{32} = \dfrac{7}{4}$. Indeed $g(x) = (4x - 7)^2$, so the parabola opens upward and touches the axis at that single point.

For $h$, $a = -2$, $b = 12$ and $c = -21$, and the two negatives multiplying together are what to watch.

$\Delta = 12^2 - 4(-2)(-21) = 144 - 168 = -24$

So $h$ has no real roots either, but this time $a = -2 < 0$, so its parabola opens downward.

Each quadratic has already been classified, so read the picture off rather than solving anything.

For $f$: no real roots means the parabola never meets the axis, and a quadratic changes sign only at a root, so $f$ keeps one sign for every real input. Since $a > 0$ the arms rise, so that one sign is positive, and completing the square confirms it.

$f(x) = 3(x - 1)^2 + 4 \ge 4 > 0$

So $f(x) > 0$ holds for every real $x$, giving $\mathbb{R}$, while $f(x) \le 0$ holds nowhere, giving $\varnothing$.

For $g$: a repeated root is not a sign change. Since $g(x) = (4x - 7)^2$ is a square it is never negative, and it is zero only where its base is, at $x = \tfrac{7}{4}$; everywhere else it is positive. So $g(x) > 0$ removes that single point and keeps all the rest, $\left(-\infty, \tfrac{7}{4}\right) \cup \left(\tfrac{7}{4}, \infty\right)$, and $g(x) \le 0$ holds only where the square is exactly zero, the one-element set $\left\{\tfrac{7}{4}\right\}$. The point is punctured out, not turned into an interval where $g$ is negative.

For $h$: no real roots again, so one sign throughout again, but now $a < 0$ and the arms fall.

$h(x) = -2(x - 3)^2 - 3 \le -3 < 0$

So $h(x) > 0$ gives $\varnothing$ and $h(x) \le 0$ gives $\mathbb{R}$. Notice that $f$ and $h$ share a negative discriminant and answer the two inequalities in opposite ways: the discriminant says there is only one sign, and the leading coefficient says which sign it is.

Test the claim against the three quadratics rather than against a memory of the rule.

Take $g$ first. Its solution set for $g(x) < 0$ is empty, since a square is never negative. Yet $g$ does have a real root, the repeated root $\tfrac{7}{4}$, where it is exactly zero.

$g\left(\tfrac{7}{4}\right) = 0, \qquad g(x) > 0 \ \text{ for every other } x$

So an empty solution set arrived with a real root present, and the claim already fails.

Now take $h$, which turns the claim around. It has no real roots at all, so the claim's conclusion holds for it; yet $h(x) < 0$ is satisfied by every real number, which is as far from empty as a solution set can get. Having no real roots does not produce an empty solution set either.

What is actually true names three things at once. The inequality $q(x) < 0$ has no solutions exactly when $q(x) \ge 0$ for every real $x$, which happens exactly when the parabola opens upward and never crosses the axis: $a > 0$ and $\Delta \le 0$. Both hypotheses are doing work. Keep $\Delta \le 0$ but let $a < 0$ and you get $h$, whose $h(x) < 0$ is all of $\mathbb{R}$; keep $a > 0$ but let $\Delta > 0$ and the parabola dips below the axis between its two roots. And the condition is $\Delta \le 0$ rather than $\Delta < 0$, because the repeated-root case still leaves the quadratic nowhere negative.

Change the symbol and the conditions change with it. For the same $q$, the inequality $q(x) \le 0$ has no solutions only when $a > 0$ and $\Delta < 0$ strictly, since a repeated root supplies one solution of $q(x) \le 0$. Emptiness is a fact about the leading coefficient, the discriminant and the symbol together, never about the discriminant on its own.

Profit is the margin on one item multiplied by the number of items sold. The margin is the price minus the buying-in cost, $x - 10$ dollars, and the number sold is $24 - x$.

$P(x) = (x - 10)(24 - x) = -x^2 + 34x - 240$

The expansion is worth doing slowly: $24x - x^2 - 240 + 10x$ collects to $-x^2 + 34x - 240$. The leading coefficient is negative, which the product form had already announced by carrying $x$ with a minus sign inside its second factor.

Now the sign question. The model arrives factored, and that is the form which makes a sign question free: the roots are $10$ and $24$, and they cut the line into three pieces.

$(x - 10)(24 - x) > 0 \iff 10 < x < 24$

Below $10$ the first factor is negative and the second positive, so the product is negative. Above $24$ the first is positive and the second negative, so the product is negative again. Only between them do the two agree in sign. The graph says the same thing without any factor table: the leading coefficient is $-1 < 0$, so the parabola opens downward and is positive between its roots.

The symbol is strict, so $10$ and $24$ are excluded, and they should be. At a price of $10$ dollars the stall is selling at cost, and at $24$ dollars it sells nothing, so the profit at each is $0$ rather than positive. The answer is the open interval $(10, 24)$.

The condition is $P(x) \ge 45$, and the first move is the one the product form tempts you out of: a product can be split factor by factor against $0$ and against nothing else, so $45$ has to be brought across first.

$-x^2 + 34x - 240 \ge 45 \iff -x^2 + 34x - 285 \ge 0$

Make the leading coefficient positive, remembering that multiplying an inequality by $-1$ reverses its symbol.

$x^2 - 34x + 285 \le 0$

Two numbers with product $285$ and sum $34$ are $15$ and $19$, so the left side is $(x - 15)(x - 19)$. The parabola now opens upward, so it is at or below zero exactly on the closed stretch between its roots, and the non-strict symbol keeps both roots, where the profit is exactly $45$ dollars.

$15 \le x \le 19$

So the target is met for prices from $15$ to $19$ dollars inclusive. That interval sits inside the one from part A, as it must: a price earning at least $45$ dollars certainly earns more than nothing.

Take the negative leading coefficient out of the first two terms before completing the square, and leave the constant outside.

$P(x) = -x^2 + 34x - 240 = -\left(x^2 - 34x\right) - 240$

Half of $34$ is $17$, and $17^2 = 289$, so add and subtract $289$ inside the bracket.

$P(x) = -\left((x - 17)^2 - 289\right) - 240 = -(x - 17)^2 + 289 - 240$

$P(x) = -(x - 17)^2 + 49$

A square is never negative and here it is multiplied by $-1$, so $-(x - 17)^2 \le 0$, with equality only at $x = 17$. Therefore $P(x) \le 49$ at every price, and $P(17) = 49$. The greatest daily profit the model allows is $49$ dollars, reached at a price of $17$ dollars.

Notice that $17$ is the midpoint of $10$ and $24$, and also the midpoint of $15$ and $19$. That is no accident: both intervals were cut from the same parabola, and a parabola is symmetric about its vertex.

Vertex form makes the whole family readable at once. The target is $P(x) \ge t$.

$-(x - 17)^2 + 49 \ge t \iff (x - 17)^2 \le 49 - t$

The question has become: how large is the square allowed to be? A square is never negative, so everything turns on the sign of $49 - t$.

While $t < 49$ the right side is positive. Taking square roots, and remembering that $\sqrt{u^2} = |u|$ rather than $u$, turns the condition into $|x - 17| \le \sqrt{49 - t}$, which says exactly that $x$ lies within $\sqrt{49 - t}$ of $17$.

$17 - \sqrt{49 - t} \le x \le 17 + \sqrt{49 - t}$

That is a closed interval, centred at $17$ whatever the target is, and it narrows as $t$ rises because $\sqrt{49 - t}$ shrinks. At $t = 0$ it is $[10, 24]$, the endpoint-inclusive counterpart of part A's $(10, 24)$, since a target of $0$ dollars accepts the two break-even prices that a strictly positive profit rejects; and at $t = 45$ it is $[15, 19]$, exactly what part B found. At $t = 49$ the half-width is $0$ and the interval collapses to the single price $17$ dollars, the one point where the maximum is reached.

Beyond that the set has no members, and it is worth saying why with all three ingredients rather than one. Rearranged as in part B, the target reads $x^2 - 34x + 240 + t \le 0$, whose leading coefficient is $1 > 0$, with discriminant

$\Delta = (-34)^2 - 4(240 + t) = 196 - 4t.$

For $t > 49$ that is negative, so this quadratic has no real roots, and since it opens upward it is strictly positive at every price; a non-strict $\le 0$ therefore has nothing to accept. All three facts are load-bearing. The same negative discriminant with a downward parabola would have made the inequality true everywhere instead of nowhere, and at $t = 49$, where the discriminant is exactly $0$, the non-strict symbol still admits the single price $17$ dollars while a strict $< 0$ would admit none.

Read back to the stall, this says the model has a ceiling. No price earns more than $49$ dollars a day, so a target above that is not a demanding target but an impossible one.

Factor first. The product of the outer coefficients is $2 \cdot (-12) = -24$ and the middle coefficient is $-5$, so split $-5x$ as $3x - 8x$.

$2x^2 - 5x - 12 = (2x + 3)(x - 4)$

Each factor supplies its own root: $2x + 3 = 0$ at $x = -\tfrac{3}{2}$, and $x - 4 = 0$ at $x = 4$. The leading coefficient is $2 > 0$, so the parabola opens upward, which puts it below the axis between the roots and above the axis outside them.

$f(x) \le 0 \iff -\tfrac{3}{2} \le x \le 4$

The symbol is non-strict, so both roots belong: there $f$ is exactly $0$, which $\le 0$ accepts. That is the closed interval $\left[-\tfrac{3}{2}, 4\right]$.

For $f(x) > 0$ take the two outside pieces instead, and this time drop the roots, since $0$ is not greater than $0$. The solution set is $\left(-\infty, -\tfrac{3}{2}\right) \cup (4, \infty)$, a union of two rays.

The proof is about real numbers, not about parabolas.

Fix a real number $x$ and look at the single real number $f(x)$. By the trichotomy of order, exactly one of $f(x) > 0$, $f(x) = 0$ and $f(x) < 0$ is true: at least one, because any real number is comparable with $0$, and at most one, because those three cases exclude each other.

$f(x) > 0 \quad \text{or} \quad f(x) = 0 \quad \text{or} \quad f(x) < 0$

Now read the two inequalities against those three cases. The statement $f(x) > 0$ is the first case alone. The statement $f(x) \le 0$ is the second and third cases together. So the three cases are shared out between the two inequalities with nothing counted twice and nothing left over: exactly one of the two holds at $x$.

Since $x$ was an arbitrary real number, no real number lies in both solution sets and no real number lies outside both. The two sets are complements of each other in $\mathbb{R}$.

Check that against part A. The union of $\left[-\tfrac{3}{2}, 4\right]$ with $\left(-\infty, -\tfrac{3}{2}\right) \cup (4, \infty)$ is the whole line, and the two share no point, precisely because each root is closed on one side and open on the other.

Nothing in the argument mentioned $a$, $b$, $c$ or $\Delta$. It used only that $f(x)$ is a real number for each real $x$. So the conclusion holds for every quadratic, opening upward or downward, with two real roots, one, or none: pairing $\le$ with $>$, or $\ge$ with $<$, always splits the line in that way, even when one of the two sets turns out to be everything and the other nothing.

The two sets are still disjoint, for the same trichotomy reason: no number is both positive and negative. What has changed is the covering.

$f(x) > 0 \ \text{ on } \ \left(-\infty, -\tfrac{3}{2}\right) \cup (4, \infty), \qquad f(x) < 0 \ \text{ on } \ \left(-\tfrac{3}{2}, 4\right)$

Both sets exclude $-\tfrac{3}{2}$ and $4$, so the union excludes them too. Those are exactly the two places where $f$ takes the one value that neither symbol accepts.

$f\left(-\tfrac{3}{2}\right) = 0, \qquad f(4) = 0$

So the union is the number line with two points punctured out, and what is missing is precisely the root set of $f$.

That identifies the general answer. For any quadratic $q$, the set where $q > 0$ and the set where $q < 0$ omit exactly the real roots of $q$, so together they cover the whole line exactly when $q$ has no real roots. The discriminant is what reports that, and only $\Delta < 0$ does it: $\Delta = 0$ leaves one root to puncture out and $\Delta > 0$ leaves two.

It is worth noticing what this does not say. When $\Delta < 0$ the two sets do cover the line, but only because one of them is all of $\mathbb{R}$ and the other is empty, the leading coefficient deciding which is which. A quadratic with no real roots never changes sign, so there is nothing there to divide.

One graph lies below another exactly where the difference of the two carries the right sign, so subtract.

$f(x) < g(x) \iff f(x) - g(x) < 0$

Subtract in the order the comparison names, $f$ minus $g$, and collect like terms.

$\left(5x^2 - 4x + 3\right) - \left(4x^2 + 2x + 1\right) = x^2 - 6x + 2$

The whole comparison has become the single quadratic inequality $x^2 - 6x + 2 < 0$. Two parabolas turned into one, which is the reason to move everything to one side before anything else happens.

Now read the coefficients: $a = 1$, $b = -6$ and $c = 2$.

$\Delta = (-6)^2 - 4(1)(2) = 36 - 8 = 28$

A positive discriminant means two distinct real roots, so the line will be cut into three pieces; $a = 1 > 0$ means the parabola opens upward, so it is negative between the roots and positive outside them; and the symbol $<$ asks for the negative part, which is the middle piece. Strictness then opens both endpoints. All of that is settled before a single root is computed, which is what the discriminant is for.