Quadratic Inequalities: Free Response

Subtract $12$ from both sides so the right side is $0$:

$x^2 + 4x \ge 12 \quad\Longrightarrow\quad x^2 + 4x - 12 \ge 0.$

Factor the left side. Two numbers multiplying to $-12$ and adding to $4$ are $6$ and $-2$:

$x^2 + 4x - 12 = (x + 6)(x - 2).$

Setting each factor to $0$ gives the roots $x = -6$ and $x = 2$.

The roots $-6$ and $2$ split the number line into three intervals: $x < -6$, $-6 < x < 2$, and $x > 2$. Test one point in each.

$x = -7:\ (-1)(-9) = 9 > 0, \qquad x = 0:\ (6)(-2) = -12 < 0, \qquad x = 3:\ (9)(1) = 9 > 0.$

The quadratic opens upward ($a = 1 > 0$), and the test points agree: positive outside the roots, negative between them. The inequality asks for $\ge 0$, and the symbol is inclusive, so the roots themselves are kept: $x \le -6$ or $x \ge 2$.

The solution $x \le -6$ or $x \ge 2$ is two separate rays, so interval notation joins them with a union:

$(-\infty, -6] \cup [2, \infty).$

The inclusive $\ge$ symbol means both roots are included, so the number line gets a CLOSED circle at $-6$ and a CLOSED circle at $2$, matching the square brackets. The shading runs outward from each circle, left from $-6$ toward $-\infty$ and right from $2$ toward $\infty$, leaving the interval between them, where the quadratic is negative, unshaded.

Check whether the trinomial fits the pattern $x^2 - 2mx + m^2 = (x-m)^2$. The middle coefficient is $-8$, so $m = 4$, and the constant term should be $m^2 = 16$, which it is:

$x^2 - 8x + 16 = (x - 4)^2.$

Setting $(x-4)^2 = 0$ gives the single repeated root $x = 4$; the discriminant confirms it, $(-8)^2 - 4(1)(16) = 64 - 64 = 0$.

The claim has two parts: a fact about squares, and what that fact implies for a scaled quadratic.

The square is never negative, and only vanishes at $x=r$. Let $x$ be any real number with $x \ne r$. Then the real number $x - r$ is itself nonzero, and a nonzero real number is either positive or negative. Squaring either case gives a positive result: a positive times a positive is positive, and a negative times a negative is also positive. So

$(x - r)^2 > 0 \quad \text{for every } x \ne r,$

while at $x = r$ the square is exactly $0$.

No interval where the sign changes. Take any nonzero constant $a$. Since $(x-r)^2$ is positive everywhere except at the single point $x = r$, the product $a(x-r)^2$ has the SAME sign as $a$ at every $x \ne r$: multiplying a positive number by $a$ never changes which sign $a$ carries. Unlike the two-distinct-root case, where crossing $r$ or $s$ flips exactly one factor negative, here there is no second factor to flip: $(x-r)$ is squared, so its own sign is erased before the product is ever formed. The curve touches the axis at $x=r$ and returns to the same side, rather than crossing to the other one.

By part B, $(x-4)^2$ is positive everywhere except at the single point $x=4$, where it equals $0$ exactly, and never negative anywhere.

$(x-4)^2 \ge 0 \text{ for every } x, \qquad (x-4)^2 = 0 \text{ only at } x = 4.$

For $(x-4)^2 \ge 0$: a square is never negative, so this holds at every real number, including $x=4$ itself where it is $0$.

For $(x-4)^2 > 0$: this excludes exactly the point where the square is $0$, so it holds everywhere except $x=4$.

For $(x-4)^2 \le 0$: since the square is never negative, the only way to satisfy $\le 0$ is for it to equal $0$ exactly, which happens only at $x=4$. There is no all-reals-or-no-solution split here the way there is with two distinct roots; the single touch point is itself the entire boundary.

The tabletop's length is $4$ cm more than its width, so with width $w$, the length is $w+4$, and the area is

$A = w(w+4) = w^2 + 4w.$

'At least $96$' becomes $A \ge 96$, and moving everything to one side compares it to $0$:

$w^2 + 4w - 96 \ge 0.$

Factor: two numbers multiplying to $-96$ and adding to $4$ are $12$ and $-8$:

$w^2 + 4w - 96 = (w+12)(w-8).$

The roots are $w=-12$ and $w=8$.

The roots $-12$ and $8$ split the line into three intervals. Testing $w=-13$: $(-1)(-21)=21>0$. Testing $w=0$: $(12)(-8)=-96<0$. Testing $w=9$: $(21)(1)=21>0$. The upward parabola is positive outside the roots, matching the test points, and the inclusive symbol keeps the endpoints:

$w \le -12 \quad \text{or} \quad w \ge 8.$

The algebra in part B produced two rays, $w \le -12$ or $w \ge 8$, because it never knew that $w$ stands for a width. A width is a physical length, so it must satisfy $w > 0$ regardless of what the inequality alone allows.

Every value in $w \le -12$ is negative, so none of them can be an actual width; that whole ray is discarded, not because the algebra was wrong, but because it answered a wider question, 'for which real numbers is $w^2+4w-96 \ge 0$', than the one being asked here.

The ray $w \ge 8$ is entirely positive, so it survives the physical constraint untouched. The set of widths that actually work is

$w \ge 8 \text{ centimeters.}$

Check each line as its own claim, in order.

The first line is simply the inequality as given. The second multiplies both sides by $-1$, which is a legal move, but multiplying (or dividing) an inequality by a NEGATIVE number reverses its direction, a rule already established for linear inequalities. The symbol here was left as $\ge$ instead of being flipped to $\le$, so that line should have read

$x^2 - 5x + 4 \le 0.$

Every line after it inherits the wrong direction, even though the factoring and the sign-pattern reasoning that follow are both individually correct.

Plug each value into the ORIGINAL inequality, not the rearranged form, since that is the one actually being solved.

At $x = 2$:

$-(2)^2 + 5(2) - 4 = -4 + 10 - 4 = 2 \ge 0,$

so $x = 2$ IS a solution. The claimed set, $x \le 1$ or $x \ge 4$, does not contain $2$, so it wrongly excludes it.

At $x = 5$:

$-(5)^2 + 5(5) - 4 = -25 + 25 - 4 = -4,$

which is not $\ge 0$, so $x = 5$ is NOT a solution. The claimed set does contain $5$ (it is $\ge 4$), so it wrongly includes it.

Both checks land inside the corrected set $1 \le x \le 4$ instead: $2$ is inside it, and $5$ is outside it. The corrected line, not the original conclusion, matches direct substitution.

There are two things to establish: why a direct check is trustworthy, and what exactly the flipped-but-not-flipped symbol did.

Why the check is trustworthy. Substituting a specific number into the ORIGINAL inequality asks one question only: is this particular numerical statement true? That question does not pass through any of the algebra in between, so however many steps came before, a wrong one among them cannot make a true statement about the original false, or a false one true. The check answers the original question directly, every time.

What the mistake actually did. Multiplying $-x^2+5x-4 \ge 0$ by $-1$ produces $x^2-5x+4$ on the other side, but the DIRECTION has to reverse along with it:

$-x^2+5x-4 \ge 0 \quad\Longleftrightarrow\quad x^2-5x+4 \le 0.$

The chain left the symbol as $\ge$, so the reasoning from that point on correctly solved $x^2-5x+4 \ge 0$, a genuinely different inequality from the one given. Every later line is valid algebra for that different problem; it was simply never the problem being asked.

For an upward-opening quadratic in $x$ (leading coefficient $1>0$), being positive for every real $x$ means never touching or crossing the axis, which happens exactly when its discriminant is negative.

With $a=1$, $b=-2m$, $c=m+6$:

$b^2-4ac = (-2m)^2 - 4(1)(m+6) = 4m^2 - 4m - 24 = 4\left(m^2 - m - 6\right).$

Requiring this negative, and dividing by the positive constant $4$ (which does not flip the direction):

$4(m^2-m-6) < 0 \quad\Longrightarrow\quad m^2 - m - 6 < 0.$

Factor to find the roots in $m$: two numbers multiplying to $-6$ and adding to $-1$ are $-3$ and $2$:

$m^2-m-6 = (m-3)(m+2).$

The roots are $m=3$ and $m=-2$. This is an upward-opening quadratic IN $m$, so it is negative between its roots, and the symbol is strict:

$-2 < m < 3.$

At $m=3$, substitute directly:

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

This is a perfect square with the single repeated root $x=3$, exactly the boundary case where the discriminant is $0$ rather than negative. A square is never negative, so $(x-3)^2 \ge 0$ for every $x$, but at $x=3$ it equals $0$ exactly, not more than $0$.

The original requirement was that the quadratic stay strictly ABOVE the axis for every $x$, and at $x=3$ it sits ON the axis instead. So $m=3$ fails the requirement by exactly one point, which is why part B's strict inequality correctly leaves it out; the same argument applies at the other boundary, $m=-2$, giving $x^2+4x+4=(x+2)^2$ touching zero at $x=-2$.