The Quadratic Formula and the Discriminant: Free Response

The equation is already in standard form, so take each coefficient complete with its sign: $a = 5$, $b = -8$, $c = 1$. The sign of $b$ is used twice, once in $-b$ and once in $b^2$, so substitute in parentheses and let the arithmetic look after it.

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(1)}}{2(5)} = \frac{8 \pm \sqrt{44}}{10}$

The discriminant is $64 - 20 = 44$: positive, and, the coefficients here being integers, not a perfect square, so the roots are irrational and an integer factoring search was never going to find them.

Simplify the radical by pulling out the largest perfect-square factor, using $44 = 4 \cdot 11$:

$\sqrt{44} = 2\sqrt{11}, \qquad x = \frac{8 \pm 2\sqrt{11}}{10}$

Every term of the numerator and the denominator carries a factor of $2$, so the whole fraction reduces:

$x = \frac{2\left(4 \pm \sqrt{11}\right)}{10} = \frac{4 \pm \sqrt{11}}{5}$

Nothing further comes out: $11$ is prime, and $4$, $\sqrt{11}$ and $5$ share no factor. The roots are $\frac{4 + \sqrt{11}}{5} \approx 1.46$ and $\frac{4 - \sqrt{11}}{5} \approx 0.14$.

The first lesson of this chapter located the axis from the coefficients alone, with no roots involved:

$x = -\frac{b}{2a} = -\frac{-8}{2(5)} = \frac{4}{5}$

In the split form of the formula the radical term is the only thing added to that centre, so its size is the distance from the axis out to a root. With $\Delta = 44$:

$\frac{\sqrt{\Delta}}{2\lvert a \rvert} = \frac{\sqrt{44}}{10} = \frac{2\sqrt{11}}{10} = \frac{\sqrt{11}}{5}$

So the axis is $x = \frac{4}{5}$ and each root sits $\frac{\sqrt{11}}{5}$ away from it, one on each side. Neither line needed a root, which is the point: the coefficients fix the centre and the discriminant fixes the spread, separately.

Split a root of $ax^2 + bx + c = 0$, with $\Delta = b^2 - 4ac > 0$, into the two pieces the formula offers:

$x = -\frac{b}{2a} \pm \frac{\sqrt{\Delta}}{2a}$

The first piece contains no sign choice. It is built from $a$ and $b$ alone, so it is the SAME number for both roots: whichever sign is taken, both roots are measured from it.

The second piece is the only place the two roots can differ, because it is the only term carrying the $\pm$. One root takes it added and the other takes it subtracted, and its size is $\frac{\sqrt{\Delta}}{2\lvert a \rvert}$ either way, since $\Delta > 0$ makes the radical a positive real number.

So the two roots are one common number, plus and minus one common distance. That already fixes their midpoint: stepping out by $\frac{\sqrt{\Delta}}{2\lvert a \rvert}$ on one side and by the same amount on the other leaves the point you started from halfway between, and the point started from is $-\frac{b}{2a}$. Their midpoint is therefore exactly the axis of symmetry, so the roots are a mirror pair straddling it, each at distance $\frac{\sqrt{\Delta}}{2\lvert a \rvert}$ on its own side.

It could not have come out otherwise. The formula offers exactly one term that tells the two roots apart, and that term appears with equal size and opposite signs, so nothing is available to push one root further from the centre than the other. This is the algebra agreeing with the picture the first lesson of the chapter gave, where the two points at which a parabola meets the $x$-axis are reflections of each other in its axis.

Since $D \ge 0$, the symbol $\sqrt{D}$ names a real number and $\left(\sqrt{D}\right)^2 = D$. Square the whole numerator, keeping its middle term:

$\left(-b + \sqrt{D}\right)^2 = b^2 - 2b\sqrt{D} + D$

Dividing by $4a^2$ and multiplying by $a$ gives the first term over the denominator $4a$:

$a\,x_+^2 = \frac{b^2 - 2b\sqrt{D} + D}{4a}$

Put the middle term over the same denominator:

$b\,x_+ = \frac{-b^2 + b\sqrt{D}}{2a} = \frac{-2b^2 + 2b\sqrt{D}}{4a}$

and the constant as well, $c = \frac{4ac}{4a}$. Now add the three numerators:

$b^2 - 2b\sqrt{D} + D - 2b^2 + 2b\sqrt{D} + 4ac$

The two radical terms are $-2b\sqrt{D}$ and $+2b\sqrt{D}$, equal in size and opposite in sign, so they cancel, and no other term contains $\sqrt{D}$. What is left is

$-b^2 + D + 4ac = -b^2 + \left(b^2 - 4ac\right) + 4ac = 0$

So $a\,x_+^2 + b\,x_+ + c = \frac{0}{4a} = 0$, and $x_+$ satisfies the equation.

Nothing was assumed about $a$, $b$ and $c$ beyond $a \ne 0$ and $D \ge 0$, so this single calculation covers every quadratic at once rather than one example at a time. Exactly what that generality buys, and what it does not, is part B's question.

A substitution check begins with a named number and confirms that the equation holds there. That establishes something real and worth having: the two values the formula produces are genuinely solutions, so the formula never invents an answer.

The gap is easy to state. Confirming that two named numbers are roots would still be true if some third number were also a solution, and the check has no way of noticing it, because it never looks at any number it was not handed. A claim about ALL solutions is a claim about numbers nobody named, so it needs an argument that starts from an ARBITRARY solution rather than from a chosen one.

The derivation is exactly that argument, and it runs in the direction the check cannot. Take any real $x$ with $ax^2 + bx + c = 0$. Dividing by $a \ne 0$, moving the constant, adding the same quantity to both sides and factoring the left side are all reversible moves that leave the solution set untouched, so that same $x$ satisfies

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

A real square equals a given non-negative number for exactly two values of its base when that number is positive, and one when it is zero, so $x + \frac{b}{2a}$ can only be $\pm\frac{\sqrt{D}}{2a}$ and $x$ can only be one of the two values the formula names. No third solution can exist.

Put the halves together and the theorem is complete: the substitution check says the formula's values really are solutions, and the reversibility of the derivation says there are no others. Both directions are needed, and neither one implies the other.

Walk back through part A and ask which step could fail. The coefficients $a$, $b$ and $c$ are only added and multiplied, and $a \ne 0$ is what allows the denominator. Exactly one step of the algebra depends on the sign of $D$: the moment the squared numerator is expanded and $\left(\sqrt{D}\right)^2$ is replaced by $D$.

$\left(-b + \sqrt{D}\right)^2 = b^2 - 2b\sqrt{D} + \left(\sqrt{D}\right)^2 = b^2 - 2b\sqrt{D} + D$

That replacement is available only when $\sqrt{D}$ is a real number, which is what $D \ge 0$ buys. And if $D < 0$ the difficulty is more basic still: there is no real $\sqrt{D}$ to write down, so $x_+$ never gets defined and there is nothing to substitute. The check does not fail; it never begins.

The honest report for $D < 0$ is the one the completed-square line already gives. That line reads

$\left(x + \frac{b}{2a}\right)^2 = \frac{D}{4a^2}$

and $4a^2 > 0$, so the right side is negative while the left side is a real square. No real square is negative, so no real $x$ satisfies the equation, and the answer is that it has no real solutions.

That is a complete answer over the real numbers rather than an unfinished one. Do not manufacture a real square root for a negative number: there is none. Read against the graph, the same sign says the parabola misses the $x$-axis entirely: its vertex sits at height $-\frac{D}{4a}$, which is positive when $a > 0$ and negative when $a < 0$, so the whole curve stays on one side of the axis, the vertex being the point that comes closest to it.

The model gives the temperature at time $t$, so "the reading is exactly $38$ degrees" is the statement that $T(t) = 38$:

$-t^2 + 12t + 15 = 38$

Collect every term onto one side against zero. Moving the left side across rather than the right turns the leading coefficient positive, which keeps the later arithmetic tidier:

$0 = t^2 - 12t + 38 - 15 = t^2 - 12t + 23$

So the equation is $t^2 - 12t + 23 = 0$. One unknown and one equation, because the model already ties the temperature to the single quantity $t$.

Rebuild the equation from the stem if you need it: $-t^2 + 12t + 15 = 38$ collects to $t^2 - 12t + 23 = 0$.

Read off $a = 1$, $b = -12$, $c = 23$, and look at the discriminant before choosing how to finish:

$\Delta = (-12)^2 - 4(1)(23) = 144 - 92 = 52$

Positive, so there are two such times. With integer coefficients in place, $52$ is not a perfect square, so those times are irrational: no factoring over the integers is going to produce them, and the formula is the tool that will.

$t = \frac{12 \pm \sqrt{52}}{2}$

Simplify the radical by its largest square factor, using $52 = 4 \cdot 13$:

$\sqrt{52} = 2\sqrt{13}, \qquad t = \frac{12 \pm 2\sqrt{13}}{2} = 6 \pm \sqrt{13}$

Both times lie inside the twelve-hour window, since $\sqrt{13}$ is between $3$ and $4$. Numerically $\sqrt{13} \approx 3.606$, so the reading passes $38$ degrees on the way up about $2.4$ hours after sunrise, and again on the way down about $9.6$ hours after sunrise.

Nothing here needs solving. Only the target changed, so only the constant term changes:

$-t^2 + 12t + 15 = 58 \ \Longrightarrow \ t^2 - 12t + 43 = 0$

and one number counts the real solutions:

$\Delta = (-12)^2 - 4(1)(43) = 144 - 172 = -28$

The discriminant is negative, so this equation has no real solutions. A time is a real number, so there is no time whatever, inside the window or outside it, at which this model reads $58$ degrees. The crop is safe on this morning, and the discriminant reported it before a single root was attempted, which is the whole reason for consulting it first.

The report stops there. A negative discriminant is a finished answer over the real numbers, not a calculation waiting on the square root of a negative number, because no such real square root exists. Read against the graph, the same fact says the horizontal line at $58$ degrees and the curve of the model never meet.

Do the same work once with the target left as a letter. Writing $M$ for the reading in degrees, setting $T(t) = M$ and collecting gives

$t^2 - 12t + (M - 15) = 0, \qquad \Delta = 144 - 4(M - 15) = 204 - 4M$

Take $M \ge 15$, the reading the model gives at sunrise, so that a time inside the window is what is at stake. For such an $M$ the model produces the reading exactly when this equation has a real solution, and the discriminant settles that in both directions:

$204 - 4M \ge 0 \ \Longleftrightarrow \ M \le 51$

If $15 \le M \le 51$ the discriminant is non-negative, so a real $t$ exists, and it lands where it is wanted. The two solutions are $t = 6 \pm \sqrt{51 - M}$, and $M \ge 15$ gives $51 - M \le 36$, so the offset is at most $6$ and both times sit inside $0 \le t \le 12$. If instead $M > 51$ the discriminant is negative, so no real $t$ exists at all and the reading never occurs. No trial is needed, and no set of trials could settle it, since trials can only ever confirm the readings that do occur.

So the highest reading is $51$ degrees, the boundary case $\Delta = 0$. There the radical vanishes and the two times coincide:

$t = \frac{12}{2} = 6$

The model peaks at $51$ degrees six hours after sunrise, and the $58$ of part C is above that ceiling, as the negative discriminant there had already reported.

The restriction to $M \ge 15$ is doing real work here. A target below the sunrise reading passes the discriminant test just as happily and fails the second line instead: the offset $\sqrt{51 - M}$ would exceed $6$, putting the earlier of the two times before sunrise, outside the window this model is used on.

Read the coefficients with their signs: $a = 8$, $b = 2$, $c = -15$. Because $c$ is negative, the $-4ac$ term is positive and ADDS:

$\Delta = 2^2 - 4(8)(-15) = 4 + 480 = 484$

Now separate the two readings. The SIGN is positive, so $8x^2 + 2x - 15 = 0$ has two distinct real roots; this holds for any real coefficients and needs no further hypothesis. Separately, $484 = 22^2$ is a perfect square AND the coefficients here are integers, so the perfect-square test applies and predicts a rational factorization before one has been found.

The formula confirms it, and the roots are rational as promised:

$x = \frac{-2 \pm 22}{16}, \qquad x = \frac{5}{4} \ \text{ or } \ x = -\frac{3}{2}$

Cash each root in for a factor and clear the fractions into the leading coefficient:

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

Multiplying back gives $8x^2 + 12x - 10x - 15 = 8x^2 + 2x - 15$, so the factorization is right.

A refutation has to meet the claim's hypothesis, or it tests nothing. So the discriminant must genuinely be positive, and the coefficients must be integers, since otherwise the perfect-square test is not even in play.

Take $x^2 - 3$, with $a = 1$, $b = 0$, $c = -3$:

$\Delta = 0^2 - 4(1)(-3) = 12 > 0$

The hypothesis holds, and there are two distinct real roots, $x = \pm\sqrt{3}$. Over the real numbers it does factor:

$x^2 - 3 = \left(x - \sqrt{3}\right)\left(x + \sqrt{3}\right)$

Over the rationals it does not. Its coefficients are integers and $12$ is not a perfect square, so the perfect-square test, hypothesis satisfied, denies a rational factorization outright. The direct reason is the same one: a rational factorization would hand over a rational root, and the only roots are $\pm\sqrt{3}$, which is irrational.

So the example has real roots and lacks a rational factorization, and the classmate's rule collapses. What a positive discriminant promises is real roots, and with them a factorization over the REALS, since each real zero buys a real factor. A rational factorization needs the strictly stronger condition that the discriminant is a perfect square, on top of rational coefficients.

First confirm the discriminant, since the claim under test is about a perfect square. With $a = \sqrt{2}$, $b = 3$, $c = \sqrt{2}$:

$\Delta = 3^2 - 4\left(\sqrt{2}\right)\left(\sqrt{2}\right) = 9 - 8 = 1$

and $1 = 1^2$ is as perfect a square as they come. If the test applied here, it would promise a rational root.

There is none. Suppose $r$ were a rational number with

$\sqrt{2}\,r^2 + 3r + \sqrt{2} = 0$

Collect the two terms carrying $\sqrt{2}$ and move the other across:

$\sqrt{2}\left(r^2 + 1\right) = -3r$

Now $r^2 + 1 \ge 1$, so it is never zero and may be divided by:

$\sqrt{2} = \frac{-3r}{r^2 + 1}$

The right-hand side is built from the rational number $r$ by multiplying, adding and dividing, so it is rational. That would make $\sqrt{2}$ rational, and it is not. The supposition is impossible, so no rational number is a root.

The roots are real, all the same, and simply irrational. One of them is $x = -\sqrt{2}$, since

$\sqrt{2}\left(-\sqrt{2}\right)^2 + 3\left(-\sqrt{2}\right) + \sqrt{2} = 2\sqrt{2} - 3\sqrt{2} + \sqrt{2} = 0$

What failed is the hypothesis, and it is worth seeing exactly where. The test is proved by a chain of equivalences: a quadratic with rational coefficients factors over the rationals exactly when it has a rational root, which happens exactly when $\sqrt{\Delta}$ is rational, because the roots are $\frac{-b \pm \sqrt{\Delta}}{2a}$ and $a$ and $b$ are rational. That last step is where rational coefficients are spent, and here $a = \sqrt{2}$ is not rational, so the chain breaks at its first link. State the hypothesis every time the test is used: without it the test is not merely weaker, it is false.

A shared point has one $x$ giving the same $y$ on both graphs, so set the two expressions equal:

$x^2 - 6x + 14 = 2x + 2$

Collect everything on one side against zero:

$x^2 - 8x + 12 = 0$

Glance at the discriminant before choosing a method:

$\Delta = (-8)^2 - 4(1)(12) = 64 - 48 = 16$

Positive and a perfect square with integer coefficients, so two rational solutions are coming and either factoring or the formula will land cleanly:

$x = \frac{8 \pm 4}{2}, \qquad x = 6 \ \text{ or } \ x = 2$

Each $x$ needs its $y$. From the line, $y = 2(2) + 2 = 6$ and $y = 2(6) + 2 = 14$, and the parabola agrees at both, since $2^2 - 6(2) + 14 = 6$ and $6^2 - 6(6) + 14 = 14$.

The graphs meet at $(2, 6)$ and $(6, 14)$.

Do the same work with $c$ carried through as a letter. Setting the two expressions equal and collecting:

$x^2 - 6x + 14 = 2x + c \ \Longrightarrow \ x^2 - 8x + (14 - c) = 0$

The number of meeting points is the number of real solutions of that equation, so "exactly one meeting point" is exactly the condition $\Delta = 0$:

$\Delta = (-8)^2 - 4(1)(14 - c) = 64 - 56 + 4c = 8 + 4c$

$8 + 4c = 0 \ \Longrightarrow \ c = -2$

At that value the equation reads $x^2 - 8x + 16 = 0$, a perfect square with the double root $x = 4$; the formula says the same thing, collapsing to $x = -\frac{b}{2a} = 4$ once the radical vanishes.

The height follows from the line, $y = 2(4) - 2 = 6$, and the parabola agrees: $4^2 - 6(4) + 14 = 6$. So $y = 2x - 2$ touches the parabola at the single point $(4, 6)$, which is the tangency case: one point, not a crossing.

A claim about every member of a family falls to a single member, so produce one.

Take $c = -5$. Setting the expressions equal and collecting:

$x^2 - 6x + 14 = 2x - 5 \ \Longrightarrow \ x^2 - 8x + 19 = 0$

One number decides the count:

$\Delta = (-8)^2 - 4(1)(19) = 64 - 76 = -12$

The discriminant is negative, so this equation has no real solutions. There is no real $x$ at which the two graphs agree, so the line $y = 2x - 5$ and the parabola have no point in common whatever, and the classmate's claim is false.

The argument behind the claim is worth naming, because the picture is doing the reasoning. An upward parabola does eventually outrun any straight line, so the line cannot stay clear of the curve on the high side forever. That is a statement about one side of the comparison only, and it does not force the two graphs to touch: the discriminant has just produced a member of the family that the curve never reaches.

And the negative discriminant is a complete answer here rather than an unfinished one. Over the real numbers there is no square root of $-12$ to take, so the honest report is that the two graphs do not meet. The same number read the other way is a prediction: since the discriminant of $x^2 - 8x + (14 - c)$ is $8 + 4c$, one glance at $c$ tells you in advance whether you are about to find two meeting points, one, or none.