The Quadratic Formula: Free Response

Standard form already, so read off the coefficients with their signs: $a = 2$, $b = -7$, $c = 4$.

Substitute: $-b = -(-7) = 7$ and $b^2 = (-7)^2 = 49$.

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(4)}}{2(2)} = \frac{7 \pm \sqrt{49 - 32}}{4} = \frac{7 \pm \sqrt{17}}{4}.$

$17$ is prime, so the radical does not simplify; $7$, $\sqrt{17}$, and $4$ share no common factor, so the fraction does not reduce either:

$x = \frac{7 + \sqrt{17}}{4}, \qquad x = \frac{7 - \sqrt{17}}{4}.$

Here $a = 3$, $b = 2$, $c = -2$. Since $c$ is negative, $-4ac$ turns positive and ADDS to $b^2$:

$x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-2)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 24}}{6} = \frac{-2 \pm \sqrt{28}}{6}.$

$28$ is not a perfect square but carries one: $28 = 4 \times 7$, $\sqrt{4} = 2$, so $\sqrt{28} = 2\sqrt{7}$:

$x = \frac{-2 \pm 2\sqrt{7}}{6}.$

Every term, $-2$, $2$, and $6$, shares a factor of $2$, so the fraction reduces:

$x = \frac{2\left(-1 \pm \sqrt{7}\right)}{6} = \frac{-1 \pm \sqrt{7}}{3}.$

$b$ appears in two places in the formula, and each treats its sign differently.

In the numerator's leading term, $-b$ REVERSES the sign. With $b = -7$,

$-b = -(-7) = 7,$

so a negative $b$ contributes a positive number. In the discriminant, $b^2$ DESTROYS the sign, since squaring a negative gives a positive:

$b^2 = (-7)^2 = 49.$

Suppose a student reads the coefficient as $b = 7$, dropping the minus sign. The discriminant does not notice: $7^2 = 49$ too, still $49 - 32 = 17$. Only the leading term changes, from $+7$ to $-7$:

$x = \frac{-7 \pm \sqrt{17}}{4},$

the negative of each correct root.

The cheap check: substitute a root back in, or test the sum of the roots. For $2x^2 - 7x + 4 = 0$ they must sum to $-\frac{b}{a} = \frac{7}{2}$; the flawed pair sums to $-\frac{7}{2}$ instead.

Every move below was already done on numbers in the last lesson; only the numbers are letters now.

Dividing by $a$ makes the leading coefficient $1$, which completing the square needs, and the constant crosses to the right:

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \quad\Longrightarrow\quad x^2 + \frac{b}{a}x = -\frac{c}{a}.$

Half the coefficient of $x$ is $\dfrac{b}{2a}$, and its square is $\dfrac{b^2}{4a^2}$. Add that to BOTH sides:

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}.$

The left side is now a perfect square. On the right, write $\dfrac{c}{a} = \dfrac{4ac}{4a^2}$ for a common denominator:

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

The discriminant has appeared in the numerator on its own, the first sign the derivation is going somewhere.

Take the square root of both sides. Across an equation a square root admits BOTH signs, so attach a $\pm$; the root splits over numerator and denominator:

$x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}.$

Subtract $\dfrac{b}{2a}$ from both sides to isolate $x$:

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}.$

Both terms already share the denominator $2a$, so they combine into one fraction, the formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Nothing about $a$, $b$, $c$ was assumed beyond $a \ne 0$, so one formula covers every quadratic.

The assumption. The first move divides every term by $a$, and division by zero is undefined, so the derivation silently requires

$a \ne 0.$

This costs nothing: $a = 0$ removes the $x^2$ term, leaving $bx + c = 0$, a LINEAR equation. Not a quadratic the formula fails on; simply not a quadratic.

The shortcut. Strictly, $\sqrt{4a^2} = 2\lvert a \rvert$, not $2a$, since a square root is never negative and $a$ might be. Suppose $a < 0$, so $2\lvert a \rvert = -2a$. The honest root step gives

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{-2a} = -\frac{b}{2a} \mp \frac{\sqrt{b^2 - 4ac}}{2a}.$

Only the $\pm$ becomes a $\mp$. Running through both signs produces the same TWO numbers, just swapped. Since the formula claims a pair of roots, not which sign delivers which, nothing is lost.

A repeated root is exactly $b^2 - 4ac = 0$, so translate that into a condition on $k$. With $a = 1$, $b = k$, $c = 9$:

$b^2 - 4ac = k^2 - 4(1)(9) = k^2 - 36.$

Set it to zero:

$k^2 - 36 = 0 \quad\Longrightarrow\quad k^2 = 36 \quad\Longrightarrow\quad k = \pm 6.$

So exactly two members of the family have a repeated root: $k = 6$ and $k = -6$.

When the discriminant is zero the radical vanishes, so the formula collapses to its first term:

$x = \frac{-b \pm \sqrt{0}}{2a} = -\frac{b}{2a} = -\frac{k}{2}.$

$k = 6$ gives $x = -3$; $k = -6$ gives $x = 3$.

Factoring confirms both, and shows a zero discriminant really means the quadratic is a perfect square:

$x^2 + 6x + 9 = (x + 3)^2, \qquad x^2 - 6x + 9 = (x - 3)^2.$

Each root sits where its one repeated factor vanishes.

The discriminant is $k^2 - 36$, and only its SIGN matters, so ask when it is positive and when negative.

Positive exactly when $k^2 > 36$, when $k$ is more than $6$ from zero in either direction:

$k^2 - 36 > 0 \quad\Longleftrightarrow\quad k > 6 \ \text{ or }\ k < -6.$

Those members have two distinct real roots. Negative exactly when $k^2 < 36$, when $k$ lies strictly between $-6$ and $6$:

$k^2 - 36 < 0 \quad\Longleftrightarrow\quad -6 < k < 6.$

Those members have complex conjugate roots. The equation $x^2 + 9 = 0$, at $k = 0$, is the most familiar, with roots $\pm 3i$.

The two values $k = \pm 6$ from part A are the border posts; every other $k$ falls on one side or the other.

Look at where $k$ enters. The discriminant is

$k^2 - 36,$

and $k$ appears only inside a square. Squaring is blind to sign: $(-k)^2 = k^2$ for every $k$. So the equation with middle coefficient $-k$ has EXACTLY the same discriminant as the one with $k$, whatever $k$ is.

That settles it before any solving. If $k_0$ makes the discriminant zero, so does $-k_0$, since it hands the discriminant the same number. Solutions come in pairs $\{k_0, -k_0\}$; a lone value could only appear if $k_0 = -k_0$, i.e. $k_0 = 0$, which does not zero $0 - 36$. So the answer had to be a symmetric pair, and part A produced one.

The same reasoning is why the two regions in part C came out symmetric about $k = 0$ too.

Take the lines in order and test each on its own.

Line 1 is correct. The coefficients really are $a = 3$, $b = -4$, $c = -4$, signs included, worth confirming since a coefficient error would poison everything after it.

Line 2 is the first mistake. In $-4ac$, both $a$ and $c$ go in with their signs. Since $c = -4$ is negative, $-4(3)(-4)$ has two minus signs and comes out POSITIVE:

$b^2 - 4ac = (-4)^2 - 4(3)(-4) = 16 + 48 = 64.$

The student wrote $16 - 48$, what you get if $c$ is fed in as $+4$. The discriminant is $64$, a perfect square, so the roots are real and rational, not complex.

Line 3 is the second mistake, independent of the first: it would be wrong even with the right discriminant. The student divides only $-b$ by $2a$, leaving the radical undivided. The formula puts the ENTIRE numerator over $2a$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \qquad \text{not} \qquad x = \frac{-b}{2a} \pm \sqrt{b^2 - 4ac}.$

So Line 3 should read $x = \dfrac{4 \pm \sqrt{64}}{6}$.

With the discriminant repaired to $64$, substitute, putting the whole numerator over $2a = 6$:

$x = \frac{-(-4) \pm \sqrt{64}}{2(3)} = \frac{4 \pm 8}{6}.$

Split the $\pm$ and reduce each fraction:

$x = \frac{4 + 8}{6} = \frac{12}{6} = 2, \qquad x = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}.$

Check both in the ORIGINAL equation. For $x = 2$:

$3(2)^2 - 4(2) - 4 = 12 - 8 - 4 = 0.$

For $x = -\tfrac{2}{3}$, square first:

$3\left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) - 4 = 3 \cdot \frac{4}{9} + \frac{8}{3} - 4 = \frac{4}{3} + \frac{8}{3} - 4 = 4 - 4 = 0.$

Both roots are real, which the perfect-square discriminant $64$ promised in advance.

The $2a$ under the radical is not a convention: the derivation manufactures it, and you can watch it appear.

Completing the square on $ax^2 + bx + c = 0$ leaves

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

and the right-hand denominator is $4a^2$, not $1$. Taking the square root splits it over top and bottom, and $\sqrt{4a^2} = 2a$ (the $\pm$ absorbs the sign of $a$):

$x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}.$

So the radical arrives ALREADY divided by $2a$, before $x$ is even isolated. The two terms then share a denominator, the only reason they merge into the formula's single fraction. The student's Line 3 kept the merged numerator but divided only half of it, a different expression, not a shortcut.

The cheap check. The roots of $ax^2 + bx + c = 0$ must sum to $-\frac{b}{a}$, here $\frac{4}{3}$. The student's roots sum to

$\left(\frac{2}{3} + 4i\sqrt{2}\right) + \left(\frac{2}{3} - 4i\sqrt{2}\right) = \frac{4}{3},$

which matches, so the sum alone does not catch it. Their product must be $\frac{c}{a} = -\frac{4}{3}$, but is $\left(\frac{2}{3}\right)^2 + 32 = \frac{4}{9} + 32$, nowhere near it. One line of arithmetic refutes the answer without rereading a single step.

Read off $a = 2$, $b = -4$, $c = 5$; $-b = 4$ and $b^2 = 16$:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(5)}}{2(2)} = \frac{4 \pm \sqrt{16 - 40}}{4} = \frac{4 \pm \sqrt{-24}}{4}.$

The discriminant is $-24$, negative, so the roots are complex. Convert the root of the negative before simplifying anything else: $\sqrt{-24} = i\sqrt{24} = 2i\sqrt{6}$, since $24 = 4 \times 6$. So

$x = \frac{4 \pm 2i\sqrt{6}}{4}.$

Every term shares a factor of $2$; reduce the fraction and split into real and imaginary parts:

$x = \frac{2 \pm i\sqrt{6}}{2} = 1 \pm \frac{\sqrt{6}}{2}i.$

The two roots form the conjugate pair $1 + \frac{\sqrt{6}}{2}i$ and $1 - \frac{\sqrt{6}}{2}i$: same real part, opposite imaginary part.

The check. For $2x^2 - 4x + 5 = 0$ the coefficients predict a sum of $-\frac{b}{a} = \frac{4}{2} = 2$ and a product of $\frac{c}{a} = \frac{5}{2}$. The pair delivers the sum immediately, since the imaginary parts cancel:

$\left(1 + \frac{\sqrt{6}}{2}i\right) + \left(1 - \frac{\sqrt{6}}{2}i\right) = 2.$

The product is a conjugate pair multiplied out: cross terms cancel and $i^2$ turns the subtraction into an addition:

$\left(1 + \frac{\sqrt{6}}{2}i\right)\left(1 - \frac{\sqrt{6}}{2}i\right) = 1^2 - \left(\frac{\sqrt{6}}{2}i\right)^2 = 1 - \frac{6}{4}(-1) = 1 + \frac{3}{2} = \frac{5}{2}.$

Both match, so the roots are consistent with the equation they came from.

The general claim. Write $D = b^2 - 4ac$ and suppose $D < 0$. Then $\lvert D \rvert = -D$ is positive, and $\sqrt{D} = i\sqrt{\lvert D \rvert}$, so the formula gives

$x = \frac{-b \pm i\sqrt{\lvert D \rvert}}{2a} = -\frac{b}{2a} \pm \frac{\sqrt{\lvert D \rvert}}{2a}\,i.$

Both $-\frac{b}{2a}$ and $\frac{\sqrt{\lvert D \rvert}}{2a}$ are REAL, since $a$, $b$, $c$ are real and $\lvert D \rvert > 0$. So the two roots share a real part and carry imaginary parts equal in size and opposite in sign, exactly a conjugate pair. Nothing here used part A's particular numbers, so it holds for every quadratic with real coefficients and a negative discriminant.

The three tools do not compete on one axis, so compare what each is FOR.

Factoring writes the quadratic as a product and reads the roots off the factors. Fastest when it applies, but that requires rational roots, i.e. a perfect-square discriminant. On $x^2 - 5x + 6 = 0$ it takes one line, $(x - 2)(x - 3) = 0$, roots $2$ and $3$. On $2x^2 - 4x + 5 = 0$ from part A there is nothing to find; hunting for factors wastes time.

Completing the square rewrites the equation rather than merely solving it, turning it into a squared bracket plus a constant. That makes it the tool of choice when the SHAPE matters, not just the roots, and it is the only one of the three that can prove anything about the other two: the formula is completing the square done once, in letters, as question 2 showed.

The formula never fails. No pattern to spot, no lucky factorization, no procedure to rerun; feed it three numbers, it hands back both roots, real or complex. On $2x^2 - 4x + 5 = 0$ it is plainly the right choice, and the only tool that produced

$x = 1 \pm \frac{\sqrt{6}}{2}i$

without inventing anything new.

Why the formula does not retire the other two. Not completing the square, since completing the square PRODUCES the formula: retiring it leaves the formula with nothing behind it but memory. Not factoring either, since factoring is faster wherever it applies, and it also runs in the other direction, building a quadratic from its roots, which the formula cannot do at all. A good habit: glance at the discriminant first; a perfect square invites factoring, anything else calls for the formula.