Sums and Products of Roots: Free Response

The quadratic is monic, so $b=-13$ and $c=40$. The sum of the roots is $-b$ and the product is $c$.

$r+s=-b=-(-13)=13, \qquad rs=c=40.$

Both totals come straight from the coefficients, with no factoring or solving needed.

Here $a=4$, $b=9$, and $c=-3$, and this quadratic is not monic. Divide both totals by $a$, with the extra minus sign on the sum only.

$r+s=-\frac{b}{a}=-\frac{9}{4}, \qquad rs=\frac{c}{a}=\frac{-3}{4}=-\frac{3}{4}.$

Dividing by $a$ is the one extra step this quadratic needs that part A did not.

The general formulas are $r+s=-\tfrac{b}{a}$ and $rs=\tfrac{c}{a}$. Set $a=1$, since a monic quadratic is exactly one with leading coefficient $1$, and see what the formulas literally become.

$r+s=-\frac{b}{1}=-b, \qquad rs=\frac{c}{1}=c.$

That is exactly the pair of monic formulas from part A. So the monic rule was never a separate rule from the general one: it is the same relationship, specialized to the one value of $a$ that division leaves untouched.

Find the sum and the product of the two given roots first.

$r+s=\frac{2}{5}+(-3)=\frac{2}{5}-\frac{15}{5}=-\frac{13}{5}, \qquad rs=\frac{2}{5}\cdot(-3)=-\frac{6}{5}.$

Drop those into the monic template, taking care that the coefficient of $x$ is the negative of the sum.

$x^2-(r+s)x+rs=x^2+\frac{13}{5}x-\frac{6}{5}.$

Multiply every term by $5$ to clear the fractions.

$5\left(x^2+\frac{13}{5}x-\frac{6}{5}\right)=5x^2+13x-6=0.$

The factors $(5x+2)$ and $(x-3)$ give roots $x=-\tfrac{2}{5}$ and $x=3$: flip the sign of each factor's constant term, then divide by the coefficient of $x$.

$-\frac{2}{5}+3=\frac{13}{5}, \qquad \left(-\frac{2}{5}\right)(3)=-\frac{6}{5}.$

The product $-\tfrac{6}{5}$ matches the required product from part A, but the sum $\tfrac{13}{5}$ does not match the required $-\tfrac{13}{5}$: it has the wrong sign. A proposed pair has to pass both tests, so matching only the product is not enough, and this factorization is incorrect.

Expanding directly gives

$(5x+2)(x-3)=5x^2-15x+2x-6=5x^2-13x-6,$

which disagrees with $5x^2+13x-6$ on the middle term's sign, the same disagreement the sum-and-product check found. But reaching it took four products and a like-term collection.

The sum-and-product route reaches the same verdict from two short totals. Because $r+s$ and $rs$ are each a single number, a sign error is easy to isolate; inside a full expansion, a sign error is just one term among several and easier to lose track of.

Test each line on its own, using only what the line before it already established.

Line 1: $\tfrac23+(-4)=\tfrac23-\tfrac{12}{3}=-\tfrac{10}{3}$, matching what Line 1 claims. Correct.

Line 2: a positive number times a negative number is always negative, so $\tfrac23\times(-4)$ must come out negative:

$\frac23\times(-4)=-\frac83.$

Line 2 claims $+\tfrac83$, dropping that sign, so Line 2 is the first line that is not justified.

Lines 3 and 4 are not independently wrong: given Line 1's sum and Line 2's (incorrect) product, plugging them into the template and then scaling by $3$ are both carried out correctly. They inherit Line 2's error rather than adding a new one.

Corrected, the chain reads $rs=-\tfrac83$, template $x^2+\tfrac{10}{3}x-\tfrac83$, and scaled by $3$: $3x^2+10x-8=0$.

Substitute each root into the corrected equation $3x^2+10x-8$ in turn.

$3\left(\frac23\right)^2+10\left(\frac23\right)-8=3\cdot\frac49+\frac{20}{3}-8=\frac43+\frac{20}{3}-8=\frac{24}{3}-8=8-8=0.$

$3(-4)^2+10(-4)-8=3(16)-40-8=48-40-8=0.$

Both checks give $0$, confirming the corrected equation genuinely has both original roots.

A concrete illustration from this very chain: Line 2's mistake changes the constant that Lines 3 and 4 then use. Line 3 applies the template correctly to that number,

$x^2-(r+s)x+rs \;\longrightarrow\; x^2+\tfrac{10}{3}x+\tfrac{8}{3},$

and Line 4 scales it correctly by $3$. Neither step invents a new error: each applies its own correct rule to whatever the previous line already established.

That is exactly why the question is 'find the FIRST bad line,' not 'find every wrong line.' Once one line is flawed, every later line built honestly on top of it inherits the flaw without adding one of its own. Checking each line against only the line immediately before it, asking whether this step follows correctly from what came just before, rather than checking every line against the final answer, is what isolates the one place the trouble actually began.

For $x^2+3x-28=0$: $b=3$, $c=-28$, so

$r+s=-b=-3, \qquad rs=c=-28.$

For $3x^2+9x-84=0$: $a=3$, $b=9$, $c=-84$, so

$r+s=-\frac{b}{a}=-\frac{9}{3}=-3, \qquad rs=\frac{c}{a}=\frac{-84}{3}=-28.$

Both equations predict the identical sum $-3$ and product $-28$. Since a pair of roots is completely pinned down once its sum and product are both known, the two equations have the same two roots (in fact $4$ and $-7$), even though the equations themselves look different.

Start from the left side and factor out $k$, which is a common factor of every term.

$kx^2-k(r+s)x+krs=k\left(x^2-(r+s)x+rs\right).$

Write $E(x)=x^2-(r+s)x+rs$, so the original equation is $k\cdot E(x)=0$. A product of two numbers is zero exactly when at least one of them is zero, so $k\cdot E(x)=0$ holds precisely when $k=0$ or $E(x)=0$. Since $k$ was chosen to be nonzero, the first option is never available, and the equation reduces to exactly $E(x)=0$.

$k\cdot E(x)=0 \text{ and } k\ne0 \quad\Longleftrightarrow\quad E(x)=0.$

So $kx^2-k(r+s)x+krs=0$ and $x^2-(r+s)x+rs=0$ have exactly the same solutions, for every nonzero $k$ and whatever $r$ and $s$ happen to be. Multiplying an equation by a nonzero constant never changes which numbers satisfy it.

Part B proved that for any roots $r,s$ and any nonzero $k$,

$kx^2-k(r+s)x+krs=0 \iff x^2-(r+s)x+rs=0.$

Since $k$ can be any of infinitely many nonzero numbers ($1$, $2$, $-5$, $\tfrac12$, and so on), there are infinitely many different-looking quadratics that all share the same pair of roots $r$ and $s$. Calling any single one of them 'the' quadratic with those roots picks out one equation from an infinite family without saying which, so it is not accurate as stated.

What fixes the family down to one member is requiring $a=1$, that is, insisting the quadratic be monic. Exactly one value of $k$ makes $kx^2-k(r+s)x+krs$ monic, namely $k=1$, and that single choice is $x^2-(r+s)x+rs=0$. So 'the monic quadratic with roots $r$ and $s$' is accurate; 'the quadratic with roots $r$ and $s$,' with no such qualifier, is not.

Let the smaller-magnitude root be $t$, so the other root, being $5$ times as large, is $5t$. Their sum is given directly, so this is already a single linear equation in $t$.

$t+5t=6t=-12.$

Solve for $t$.

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

So the two roots are $t=-2$ and $5t=5(-2)=-10$. Check: $-2+(-10)=-12$, matching the given sum, and $-10$ is indeed $5$ times $-2$.

The general formulas are $r+s=-\dfrac{b}{a}$ and $rs=\dfrac{c}{a}$, with $a=2$ here. The sum was already given as $-12$, so use it directly to solve for $b$.

$-\frac{b}{2}=-12 \quad\Longrightarrow\quad b=24.$

The product of the two roots found in part A is

$rs=(-2)(-10)=20,$

and setting that equal to $c/a$ solves for $c$.

$\frac{c}{2}=20 \quad\Longrightarrow\quad c=40.$

So $b=24$ and $c=40$; the equation $2x^2+24x+40=0$ reduces to $x^2+12x+20=0$, which confirms the same two roots.

With the same setup, roots $t$ and $5t$, the product is

$t\cdot5t=5t^{2},$

which is quadratic in $t$, not linear. Solving it means dividing by $5$ and then taking a square root of what remains, which hands back two possible values of $t$, one positive and one negative, rather than one. Deciding which sign of $t$ actually fits the problem becomes an extra step.

The sum route used in part A avoided all of that: $t+5t=6t$ is already linear in $t$, so solving it is a single division with no sign to choose and no squaring involved. Whenever a ratio condition is combined with the sum, the scale factor comes out in one step; combined with the product, it takes an extra step and an extra decision.