Sum and Product of Roots: Free Response

Read the coefficients first: $a = 4$, $b = -9$, $c = -6$. Before either total means anything, check that there is a real pair for it to describe.

$\Delta = b^2 - 4ac = (-9)^2 - 4(4)(-6) = 81 + 96 = 177.$

That is positive, so the quadratic has two real roots and the factorization $a(x - r_1)(x - r_2)$ is available over the real numbers. Now read both totals off.

$r_1 + r_2 = -\frac{b}{a} = \frac{9}{4}, \qquad r_1 r_2 = \frac{c}{a} = -\frac{3}{2}.$

Only the sum takes the extra minus sign; the product keeps the sign of $c$. Since $177$ is not a perfect square the roots themselves are irrational, which is exactly the situation this method is for: no solving happened at any point.

Here $a = 6$, $b = -31$, and $c = 5$, so the two totals are

$r_1 + r_2 = \frac{31}{6}, \qquad r_1 r_2 = \frac{5}{6}.$

Subtract the known root from the sum, which peels the other one off in a single step.

$r_2 = \frac{31}{6} - 5 = \frac{31 - 30}{6} = \frac{1}{6}.$

The product is now an independent check, because none of the arithmetic above went into it.

$5 \cdot \frac{1}{6} = \frac{5}{6}.$

That is the product the coefficients demanded, so both totals agree and the second root is $\tfrac{1}{6}$. Agreement on one total alone would not have been enough: the sum and the product are two independent conditions, and the right pair has to satisfy them together.

The two forms are the same polynomial, so write one under the other. Expanding the factored form gives

$a(x - r_1)(x - r_2) = ax^2 - a(r_1 + r_2)x + a r_1 r_2.$

Two expressions that take the same value at every $x$ must carry the same coefficient on each power of $x$. The squared terms agree already, so compare the other two against $ax^2 + bx + c$:

$b = -a(r_1 + r_2), \qquad c = a r_1 r_2.$

Dividing each by $a$, which is legal because $a \ne 0$, turns them into the two totals. Notice that no root was ever computed: the totals fell out of the letters, and that is why they can be read with no solving.

The hypothesis sits in the very first line. Writing $ax^2 + bx + c$ as $a(x - r_1)(x - r_2)$ with real numbers $r_1$ and $r_2$ requires the quadratic to have real roots, and that is exactly the condition $\Delta \ge 0$. When $\Delta$ is negative there is no real pair for the totals to be about, so the discriminant check in part A is part of the method and not a formality.

Write $r_1$ and $r_2$ for the two side lengths. The drawing hands over the sum directly, $r_1 + r_2 = 21$, so what is missing is the product. The diagonal is what supplies it: it cuts the panel into two right triangles, so

$r_1^2 + r_2^2 = 15^2 = 225.$

That is the sum of the squares, and this lesson's identity connects it to the sum and the product. Rearranging $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2$ for the product,

$r_1 r_2 = \frac{(r_1 + r_2)^2 - (r_1^2 + r_2^2)}{2} = \frac{441 - 225}{2} = 108.$

Now both totals are known, so the two sides are the roots of the monic quadratic built from them.

$x^2 - 21x + 108 = (x - 9)(x - 12).$

The sides are $9$ cm and $12$ cm. Both came out positive, as side lengths have to be, and that was never in doubt: a positive sum together with a positive product cannot be achieved using a negative number, since a negative root would force the other root negative too and their sum with it. Checking back, $9 + 12 = 21$ and $81 + 144 = 225$, which is $15^2$.

Run the same two steps on the new measurements. The sum is unchanged at $s = 21$, and the diagonal now gives $r_1^2 + r_2^2 = 14^2 = 196$, so the product would have to be

$p = \frac{441 - 196}{2} = \frac{245}{2} = 122.5.$

Before attempting to find any sides, test whether a real pair with that sum and that product exists at all.

$s^2 - 4p = 441 - 4(122.5) = 441 - 490 = -49.$

A real pair with sum $s$ and product $p$ requires $s^2 - 4p \ge 0$, and here that quantity has come out negative. So no two real lengths add to $21$ and have squares totalling $196$, and the panel cannot be fabricated.

Notice what the verdict cost: nothing was solved and no factorization was attempted. The sign of a single number settled it, which is the whole point of reading a quadratic through its two totals.

In letters, the diagonal gives $r_1^2 + r_2^2 = d^2$, so the product is

$p = \frac{s^2 - d^2}{2}.$

Substituting that into the existence test collapses it to a statement about $s$ and $d$ alone:

$s^2 - 4p = s^2 - 2(s^2 - d^2) = 2d^2 - s^2.$

So the claim to prove is that such a rectangle exists exactly when $2d^2 \ge s^2$, and an "exactly when" needs both directions.

Forward, suppose the rectangle exists. Then its sides are real numbers with sum $s$ and product $p$, so

$2d^2 - s^2 = s^2 - 4p = (r_1 - r_2)^2,$

which is the square of a real number and therefore not negative. Hence $2d^2 \ge s^2$.

Conversely, suppose $2d^2 \ge s^2$. Put $p = \tfrac{s^2 - d^2}{2}$ and consider $x^2 - sx + p$. That quadratic has discriminant $s^2 - 4p = 2d^2 - s^2$, which the assumption makes nonnegative, so its two roots are real, and by the two relations they have sum $s$ and product $p$. They are both positive exactly when $p > 0$, that is when $d < s$, which any genuine rectangle satisfies, since a diagonal is shorter than the two sides laid end to end. So a rectangle with that side-sum and that diagonal exists.

Applying it back, $s = 21$ sets the bar at $2d^2 \ge 441$. The second panel had $2(14^2) = 392$, which falls $49$ short. The smallest diagonal that would work has $2d^2 = 441$, so $d^2 = 220.5$ and $d$ is about $14.85$ cm: the drawing missed by roughly $0.85$ cm. At exactly that diagonal the two roots coincide and the panel is a square.

Read the two totals from $a = 5$, $b = -8$, $c = -3$.

$r_1 + r_2 = \frac{8}{5}, \qquad r_1 r_2 = -\frac{3}{5}.$

The sum of the squares is symmetric in the roots, so rewrite it in those two before any number goes in.

$r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2.$

Now substitute.

$\left(\frac{8}{5}\right)^2 - 2\left(-\frac{3}{5}\right) = \frac{64}{25} + \frac{30}{25} = \frac{94}{25}.$

The cross term is always subtracted, and here the product is negative, so subtracting it increases the total. Stopping at $\tfrac{64}{25}$ would be squaring the sum and forgetting the correction.

The same two totals, with a different identity:

$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2.$

Substituting, and noting that four times a negative product is added,

$\frac{64}{25} + \frac{60}{25} = \frac{124}{25}.$

Now compute the discriminant of the same quadratic and divide it by $a^2$:

$\frac{\Delta}{a^2} = \frac{(-8)^2 - 4(5)(-3)}{5^2} = \frac{64 + 60}{25} = \frac{124}{25}.$

The two agree, as the identity $(r_1 - r_2)^2 = \tfrac{\Delta}{a^2}$ says they must. Read that identity from right to left and it accounts for the discriminant itself: whenever the two roots are real their squared gap cannot be negative, and $\Delta$ is $a^2$ times that gap, so real roots force $\Delta \ge 0$. Turn that around and a negative $\Delta$ rules a real pair out, which is why one number settles the question.

Swap the labels on the two roots and watch each side. The coefficients $a$, $b$, and $c$ do not move: it is the same quadratic, and the pair of roots is the same pair whichever one is called first. The difference does move:

$r_2 - r_1 = -(r_1 - r_2).$

So one labelling gives a certain value and the other gives its negative, while the coefficients report identical numbers in both cases. Anything the coefficients determine has to survive that relabelling unchanged, and $r_1 - r_2$ does not, so the coefficients cannot determine it. This is what "symmetric" means: the sum, the product, and $r_1^2 + r_2^2$ are all unchanged by the swap, and all three are fixed by $b$ and $c$.

The square is unchanged, because $(-(r_1 - r_2))^2 = (r_1 - r_2)^2$, and that is exactly why part B worked. The most the coefficients can hand over is the size of the gap,

$|r_1 - r_2| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{124}}{5} = \frac{2\sqrt{31}}{5},$

with its sign left undecided. The classmate's step from the square to the difference is the one that fails.

The monic template needs only the sum and the product of the two given roots.

$r_1 + r_2 = \frac{3}{4} + (-2) = -\frac{5}{4}.$

$r_1 r_2 = \frac{3}{4} \cdot (-2) = -\frac{3}{2}.$

Drop both into $x^2 - (r_1 + r_2)x + r_1 r_2$, remembering that the coefficient of $x$ is the negative of the sum, so a negative sum produces a positive middle coefficient.

$x^2 - \left(-\frac{5}{4}\right)x + \left(-\frac{3}{2}\right) = x^2 + \frac{5}{4}x - \frac{3}{2}.$

A quick check at $x = -2$: $4 - \tfrac{5}{2} - \tfrac{3}{2} = 0$, as required. "Monic" is what makes this a single answer rather than one of many.

Every member of that family has the same two roots, so the only thing to determine is the value of $a$ that sends the graph through $(0, 12)$. The $y$-intercept is the value at $x = 0$:

$a\left(0 - \frac{3}{4}\right)(0 + 2) = -\frac{3a}{2}.$

Set that equal to $12$, which is one linear equation in $a$.

$-\frac{3a}{2} = 12 \implies a = -8.$

Now expand with that value.

$-8\left(x - \frac{3}{4}\right)(x + 2) = -8x^2 - 10x + 12.$

At $x = 0$ this is $12$, as stipulated, and it still vanishes at $\tfrac{3}{4}$ and at $-2$, because multiplying by a nonzero constant moves no root. Note that $a$ came out negative: the stipulated $y$-intercept, not the roots, is what forced that.

Take the two quadratics already built, $x^2 + \tfrac{5}{4}x - \tfrac{3}{2}$ and $-8x^2 - 10x + 12$, and ask what each says about its graph.

Both meet the $x$-axis at $\tfrac{3}{4}$ and at $-2$, since multiplying by a nonzero constant changes no root. Both therefore have the same axis of symmetry, the vertical line through the midpoint of those two intercepts:

$x = \frac{1}{2}\left(\frac{3}{4} + (-2)\right) = -\frac{5}{8},$

which is also what $-\tfrac{b}{2a}$ returns for each of them, $-\tfrac{5/4}{2}$ in the first case and $\tfrac{10}{-16}$ in the second.

What differs is everything that depends on $a$. One opens upward and the other opens downward, so one has a minimum where the other has a maximum. Their $y$-intercepts are $-\tfrac{3}{2}$ and $12$. Their vertex heights differ accordingly, and so do their ranges.

So a pair of roots fixes where the graph meets the $x$-axis and where its mirror line sits, and leaves the leading coefficient entirely free. A request for the quadratic with those roots therefore picks out a whole family, every nonzero $a$ giving a member, and not one quadratic. It names a single quadratic only once something more is stipulated: monic, as in part A, or a required value at some point other than the two $x$-intercepts, as in part B, or a leading coefficient given outright. That last qualification matters: a required value AT an intercept is no help, since every member is already zero there and the condition collapses to $0 = 0$.

The first two coefficients fix the sum, whatever $k$ does:

$r_1 + r_2 = -\frac{-18}{3} = 6.$

The product is where $k$ enters:

$r_1 r_2 = \frac{k}{3}.$

The sum of the squares is symmetric, so it reduces to those two.

$r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2 = 36 - \frac{2k}{3}.$

The result is linear in $k$, which is the whole reason this question can be answered without ever writing down a root. This is an identity in $k$; whether any real pair of roots stands behind it is a separate question, and part C is where it gets asked.

Set the expression from part A equal to the required value and solve.

$36 - \frac{2k}{3} = 30 \implies \frac{2k}{3} = 6 \implies k = 9.$

That is not the end of the job. The identity used to build the expression assumed there was a real pair of roots to build it from, so check the discriminant at $k = 9$.

$\Delta = (-18)^2 - 4(3)(9) = 324 - 108 = 216.$

That is positive, so $3x^2 - 18x + 9$ does have two real roots and the value $30$ genuinely describes them. The roots happen to be $3 \pm \sqrt{6}$, and squaring and adding them returns $30$, though finding them was never necessary.

The roots are real exactly when the discriminant is not negative.

$\Delta = (-18)^2 - 4(3)k = 324 - 12k \ge 0 \iff k \le 27.$

On that range, look at what part A's expression can do. As $k$ increases, $36 - \tfrac{2k}{3}$ decreases, so it is smallest at the largest $k$ still allowed.

$36 - \frac{2(27)}{3} = 36 - 18 = 18.$

So $r_1^2 + r_2^2 \ge 18$ for every $k$ that leaves the roots real, and $18$ is actually reached: at $k = 27$ we have $3x^2 - 18x + 27 = 3(x - 3)^2$, the two roots coincide at $3$, and $9 + 9 = 18$.

The expression on its own cannot deliver that bound, because it is defined for every $k$ and keeps returning values below $18$ once $k$ passes $27$. At $k = 30$, for instance, it returns $16$, but there $\Delta = 324 - 360 = -36$ and the quadratic has no real roots at all, so there is no pair of real numbers for the number $16$ to be about. The bound comes from the discriminant, not from the expression, which is the same point the lesson makes when it insists that not every sum and product come from a real pair.