Introduction to Quadratics: Free Response

Move every term to the same side so the other side reads $0$. Subtract $6x$ from both sides of $x^2 + 7 = 6x$:

$x^2 - 6x + 7 = 0.$

The terms are already in order of decreasing power. Comparing to $ax^2 + bx + c = 0$ gives $a = 1$, $b = -6$, and $c = 7$.

Substitute each candidate into $x^2 - 3x - 10$ and see whether the result is $0$.

For $x = 5$:

$5^2 - 3(5) - 10 = 25 - 15 - 10 = 0.$

The result is $0$, so $x = 5$ is a root. For $x = -1$:

$(-1)^2 - 3(-1) - 10 = 1 + 3 - 10 = -6.$

That is not $0$, so $x = -1$ is not a root, even though it looked like a reasonable guess.

Set $a = 0$ in the standard form $ax^2 + bx + c = 0$ and see what survives. The squared term becomes $0 \cdot x^2 = 0$, so the equation collapses to

$bx + c = 0.$

That is an ordinary linear equation, with no squared term anywhere in it. Since the whole reason an equation is called quadratic is the presence of that squared term, an equation with $a = 0$ has lost the one feature that earns the name, no matter what $b$ and $c$ are. That is exactly why the definition insists on $a \neq 0$.

Before solving for $k$, decide which factor is even capable of being zero at $x = 9$. Evaluate the KNOWN factor first:

$9 + 6 = 15 \neq 0.$

Since $x + 6$ is not zero at $x = 9$, the factor $x - k$ must be the one that is zero there. Set it equal to $0$ and solve:

$9 - k = 0 \;\Rightarrow\; k = 9.$

With $k = 9$, the equation is $(x - 9)(x + 6) = 0$. Apply the zero-product property to each factor:

$x - 9 = 0 \quad\text{or}\quad x + 6 = 0.$

Solving each gives the two roots:

$x = 9 \qquad\text{or}\qquad x = -6.$

The root $x = 9$ is the one given in the stem; $x = -6$ is the one this part adds.

Part A worked because a product of two numbers is $0$ exactly when at least one of them is $0$. Nothing of that kind is true of $12$:

$(1)(12) = 12, \qquad (2)(6) = 12, \qquad (-3)(-4) = 12$

Three different ways, and none of the six numbers in them is forced to be anything. So from $(x - k)(x + 6) = 12$ you cannot conclude that $x - k = 12$, or that $x + 6 = 12$, or anything at all about either factor on its own. The split is legal for $0$ and for no other number.

What has to happen first is that the product is expanded and every term moved to one side, so the equation reads something $= 0$. Only then is there a product equal to zero, and only then does the factor-by-factor step become available again.

Take the square root of both sides of $x^2 = 169$, keeping BOTH signs, since $169$ is a positive perfect square:

$x = \pm\sqrt{169} = \pm 13.$

So $x = 13$ or $x = -13$; both check, since $13^2 = 169$ and $(-13)^2 = 169$.

Treat $(x - 12)$ as a single quantity and take the square root of both sides, keeping both signs:

$x - 12 = \pm\sqrt{4} = \pm 2.$

That splits into two one-step equations. Adding $12$ to both sides of each:

$x = 14 \qquad\text{or}\qquad x = 10.$

None of these three needs to be solved to answer the question; only the sign of the right-hand side matters.

$x^2 = 81: \ 81 > 0 \Rightarrow \text{two real solutions.}$

$x^2 = 0: \ 0 \text{ is neither positive nor negative} \Rightarrow \text{one real solution.}$

$x^2 = -81: \ -81 < 0 \Rightarrow \text{no real solution.}$

The single fact deciding the count in every case is whether the number on the right is positive, zero, or negative, exactly the same three-way split you get from $x^2 = k$ in general.

Isolate the squared group before touching a square root. Divide both sides of $4(x - 5)^2 = 64$ by $4$:

$(x - 5)^2 = 16.$

Now take the square root of both sides, keeping both signs:

$x - 5 = \pm 4.$

Add $5$ to both sides of each equation:

$x = 9 \qquad\text{or}\qquad x = 1.$

Isolate $(x - 2)^2$ completely before deciding anything. Subtract $20$ from both sides of $3(x - 2)^2 + 20 = 5$:

$3(x - 2)^2 = -15.$

Divide both sides by $3$:

$(x - 2)^2 = -5.$

A real number squared is never negative, so no real value of $x - 2$, and therefore no real value of $x$, can satisfy this. The equation has no real solution.

The argument in part B never showed that no number of any kind can square to $-5$; it showed only that no REAL number can. Reread exactly what part B put on the table:

$(x - 2)^2 = -5.$

Every step after that, from naming the sign of $-5$ to concluding there is no solution, worked entirely inside the real numbers, because that is the only number system this course has used so far.

So 'no solution, period' claims more than the argument delivered. The accurate statement is 'no real solution,' a claim about exactly the numbers the argument actually considered, no more and no less.

A universal claim fails the moment ANY single case breaks it, so pick a value of $k$ deliberately away from the ordinary two-solution case. Try $k = 0$:

$(x + 4)^2 = 0.$

Taking the square root of both sides, the $\pm$ collapses because $+0$ and $-0$ are the same number:

$x + 4 = 0 \;\Rightarrow\; x = -4.$

That is a SINGLE solution, not two, so $k = 0$ is a genuine counterexample to the claim as stated for every real $k$.

Choose a value of $k$ on the other side of zero from the one used in part A. Try $k = -9$:

$(x + 4)^2 = -9.$

The left side is a real number squared, so it can never be negative, no matter what $x$ is. Since $-9$ is negative, no real $x$ can satisfy this equation, and there is no real solution.

Line the two examples up against what happened to the equation.

$k = 0 \ (\text{part A}): \ \text{exactly one real solution.}$

$k = -9 \ (\text{part B}): \ \text{no real solution.}$

Only a POSITIVE value of $k$ has ever produced two distinct real solutions in this lesson: $x + 4 = \pm\sqrt{k}$ gives two different numbers exactly when $\sqrt{k}$ is a nonzero real number, which needs $k > 0$. So the exact condition is $k > 0$; $k = 0$ is the boundary where those two solutions merge into one, and every $k < 0$ is on the side with no real solution at all.