Chapter 4 — Quadratic and Nonlinear Equations

Answer: B

Explanation

Two things multiply to zero, so at least one is zero. Set each bracket to zero in turn:

$$x - 4 = 0 \quad\Longrightarrow\quad x = 4$$ $$x + 7 = 0 \quad\Longrightarrow\quad x = -7$$

So $x = 4$ or $x = -7\mathclose{.}$

Notice the sign flip: the bracket says $-4$ and the solution is $+4\mathclose{.}$ Solving $x - 4 = 0$ means adding 4 to both sides, so the sign changes. Getting this backwards is what the other three options are made of.

Check both in the original: $(4-4)(4+7) = 0 \times 11 = 0$ ✓ and $(-7-4)(-7+7) = -11 \times 0 = 0$ ✓

Why each wrong option is wrong:

• A, $-7$ and $-4$ — made both negative, flipping only one of the two signs.
• C, $4$ and $7$ — copied the numbers straight out of the brackets with no sign change at all.
• D, $-4$ and $7$ — flipped both signs but in the wrong direction.

Takeaway: A factor $(x - a)$ gives the solution $x = a$ — the sign flips. Then substitute both back; each should make one bracket exactly zero.

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Answer: B

Explanation

Find two numbers that multiply to $-14$ and add to $-5\mathclose{.}$

The constant is negative, so one number is positive and one negative. The pairs multiplying to $-14\mathclose{:}$ $1$ and $-14$ (adds to $-13)\mathclose{,}$ $-1$ and $14$ (adds to $13)\mathclose{,}$ $2$ and $-7$ (adds to $-5$ ✓).

$$x^2 - 5x - 14 = (x + 2)(x - 7) = 0$$

So $x = -2$ or $x = 7\mathclose{.}$

Check: $(-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 0$ ✓ and $49 - 35 - 14 = 0$ ✓

Why each wrong option is wrong:

• A, $-7$ and $2$ — the right pair of sizes with the signs swapped. Those come from $(x+7)(x-2) = x^2 + 5x - 14\mathclose{,}$ which has the middle term $+5x\mathclose{.}$
• C, $2$ and $7$ — both positive, which would need a $+14$ constant.
• D, $-14$ and $1$ — a pair multiplying to $-14$ that does not add to $-5\mathclose{.}$

Takeaway: Multiply to $c\mathclose{,}$ add to $b$ — both conditions, every time. When you have a candidate pair, check the sum before you write anything down.

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Answer: A

Explanation

Get zero on one side first — the factoring method needs it:

$$x^2 + 3x - 10 = 0$$

Two numbers multiplying to $-10$ and adding to $3\mathclose{:}$ that is $5$ and $-2\mathclose{.}$

$$(x + 5)(x - 2) = 0 \quad\Longrightarrow\quad x = -5 \text{ or } x = 2$$

Their sum is $-5 + 2 = -3\mathclose{.}$

There is a shortcut worth knowing. For $ax^2 + bx + c = 0\mathclose{:}$

$$\text{sum of solutions} = -\frac{b}{a} \qquad \text{product of solutions} = \frac{c}{a}$$

Here $-\frac{3}{1} = -3$ ✓ and the product would be $\frac{-10}{1} = -10\mathclose{.}$ When a question asks only for the sum or the product, this answers it without solving at all.

Why each wrong option is wrong:

• B, $3$ — the sign of the sum. It is $-\frac{b}{a}\mathclose{,}$ and $b = 3\mathclose{.}$
• C, $7$ — added the two numbers ignoring their signs.
• D, $10$ — took the 10 on the right for the product of the solutions. Moved across it becomes $-10\mathclose{,}$ and the question asks for the sum in any case.

Takeaway: Move everything to one side before factoring. And remember sum $= -\frac{b}{a}\mathclose{,}$ product $= \frac{c}{a}$ — they answer a whole family of questions in one line.

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Section 2 — The Quadratic Formula and the Discriminant

When factoring does not come within about twenty seconds, use the formula. It always works.

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

The part under the root has a name and a job of its own. The discriminant:

$$\Delta = b^2 - 4ac$$

It tells you how many real solutions there are, without solving:

Questions asking "for what value of $k$ does this have exactly one solution" are always discriminant questions. Set $b^2 - 4ac = 0$ and solve.

Two cautions on using the formula:

• $b$ carries its own sign. In $x^2 - 5x + 6\mathclose{,}$ $b = -5\mathclose{,}$ so $-b = +5\mathclose{.}$
• $-4ac$ with a negative $c$ becomes positive. In $x^2 + 2x - 8\mathclose{,}$ $-4(1)(-8) = +32\mathclose{.}$

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Answer: D

Explanation

Here $a = 2\mathclose{,}$ $b = 3\mathclose{,}$ $c = -2\mathclose{.}$ Work out the discriminant first:

$$b^2 - 4ac = 3^2 - 4(2)(-2) = 9 + 16 = 25$$

Positive, so there are two solutions — and 25 is a perfect square, so they will be tidy. Now the formula:

$$x = \frac{-3 \pm \sqrt{25}}{2(2)} = \frac{-3 \pm 5}{4}$$

$$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac12 \qquad x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$

Check the smaller one: $2(4) + 3(-2) - 2 = 8 - 6 - 2 = 0$ ✓

Note $-4ac = -4(2)(-2) = +16\mathclose{:}$ two negatives made it positive, which is what turned 9 into 25 rather than into $-7\mathclose{.}$

Why each wrong option is wrong:

• B, $- \frac{1}{2}$ and $2$ — both solutions with their signs flipped, from using $+b$ instead of $-b$ on the top.
• C, $-2$ and $2$ — solved $2x^2 - 8 = 0\mathclose{,}$ dropping the middle term.
• A, $-4$ and $1$ — divided by 2 instead of by $2a = 4\mathclose{:}$ $\frac{-3 \pm 5}{2}\mathclose{.}$ The whole bottom of the formula is $2a\mathclose{,}$ and here $a$ is not 1.

Takeaway: Compute $b^2 - 4ac$ first — it tells you how many solutions to expect and warns you if the arithmetic has gone wrong. Watch the sign of $c\mathclose{:}$ $-4ac$ with negative $c$ adds.

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Answer: 8

Explanation

Touching the $x$-axis at exactly one point means $x^2 + kx + 16 = 0$ has exactly one real solution, and that happens when the discriminant is zero:

$$b^2 - 4ac = 0$$

Here $a = 1\mathclose{,}$ $b = k\mathclose{,}$ $c = 16\mathclose{:}$

$$k^2 - 4(1)(16) = 0 \quad\Longrightarrow\quad k^2 = 64 \quad\Longrightarrow\quad k = \pm 8$$

The question says $k$ is positive, so $k = 8\mathclose{.}$

Check: $x^2 + 8x + 16 = (x + 4)^2\mathclose{,}$ which is zero only at $x = -4$ — one solution ✓

Notice what "exactly one solution" always means structurally: the quadratic is a perfect square. That gives you a second route to the same answer — the middle coefficient of a perfect square is twice the root of the constant, and $2\sqrt{16} = 8\mathclose{.}$

This is a grid-in. Enter 8.

Takeaway: One solution ⇒ discriminant zero ⇒ perfect square. All three say the same thing, and each gives you a way in. Watch for a question restricting $k$ to positive values — $k^2 = 64$ has two answers and the grid takes one.

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Answer: A

Explanation

No real solutions means the discriminant is negative:

$$b^2 - 4ac < 0$$

With $a = 1$ and $b = -6\mathclose{:}$

$$36 - 4c < 0 \quad\Longrightarrow\quad 36 < 4c \quad\Longrightarrow\quad c > 9$$

Of the options, only $c = 10$ is greater than 9.

Run the four values through $36 - 4c$ to see the whole picture:

D and C leave the discriminant positive, so each gives two solutions. B is the one to watch. $c = 9$ makes the discriminant exactly zero, which is one solution, not none. That boundary is where these questions are always set.

Takeaway: None ⇒ negative discriminant. Zero ⇒ exactly one. The option sitting exactly at the boundary is always offered, so work out which side of it you need before choosing.

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Section 3 — Vertex Form and the Turning Point

The graph of a quadratic is a parabola — a symmetric U shape. Its turning point is the vertex, and questions about maximum or minimum values are always vertex questions.

Two ways to find it:

From the formula. The vertex sits at

$$x = -\frac{b}{2a}$$

then substitute that $x$ back to get the $y$-value. This is the fastest route and it always works.

From vertex form. If the quadratic is written

$$y = a(x - h)^2 + k$$

the vertex is $(h, k)$ — read straight off. Note the sign flip on $h\mathclose{:}$ $y = (x - 3)^2 + 2$ has its vertex at $(3, 2)\mathclose{,}$ and $y = (x + 3)^2 + 2$ at $(-3, 2)\mathclose{.}$

Which way it opens, and so whether the vertex is a minimum or a maximum:

• $a > 0$ — opens upward, vertex is the minimum
• $a < 0$ — opens downward, vertex is the maximum

The parabola is symmetric about the vertical line through its vertex, so the two $x$-intercepts are always the same distance from it — which means the vertex's $x$ is the average of the two roots.

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Answer: B

Explanation

Compare with $y = a(x - h)^2 + k\mathclose{,}$ whose vertex is $(h, k)\mathclose{:}$

$$y = 2(x - \mathbf{3})^2 + \mathbf{5}$$

So $h = 3$ and $k = 5\mathclose{,}$ and the vertex is $(3, 5)\mathclose{.}$

The sign flip is the whole question. The bracket reads $x - 3\mathclose{,}$ so $h = +3\mathclose{.}$ Only $k$ is read off unchanged.

Check it directly. At $x = 3$ the squared term is zero, giving $y = 5$ — the smallest $y$ can be, since a square is never negative and $a = 2$ is positive. So $(3, 5)$ is the minimum ✓ At $x = 2$ and $x = 4\mathclose{,}$ $y = 7$ — higher on both sides ✓

Why each wrong option is wrong:

• A, $\left( -3, \ 5\right)$ — kept the minus from the bracket. $(x - 3)$ means $h = 3\mathclose{.}$
• C, $\left( 3, \ -5\right)$ — flipped $k$ as well. Only $h$ flips.
• D, $\left( 2, \ 3\right)$ — read the leading coefficient and $h$ as if they were the coordinates.

Takeaway: $y = a(x-h)^2 + k$ has vertex $(h, k)$ — the $h$ flips sign, the $k$ does not. Substituting $x = h$ confirms it in one step.

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Answer: C

Explanation

The vertex of $y = ax^2 + bx + c$ is at

$$x = -\frac{b}{2a}$$

Here $a = 1$ and $b = -8\mathclose{:}$

$$x = -\frac{-8}{2(1)} = \frac{8}{2} = 4$$

The two minus signs cancel, which is where option B goes wrong.

Check by symmetry. Factor the quadratic's roots or just test either side: at $x = 3\mathclose{,}$ $y = 9 - 24 + 3 = -12\mathclose{;}$ at $x = 4\mathclose{,}$ $y = 16 - 32 + 3 = -13\mathclose{;}$ at $x = 5\mathclose{,}$ $y = 25 - 40 + 3 = -12\mathclose{.}$ Lower at 4 than on either side, and symmetric about it ✓

Why each wrong option is wrong:

• A, $-8$ — read $b$ off the equation with no formula applied.
• B, $-4$ — divided but dropped the leading minus, so the two negatives did not cancel.
• D, $8$ — took $-b$ and forgot to divide by $2a\mathclose{.}$

Takeaway: $x = -\frac{b}{2a}\mathclose{,}$ and $b$ brings its own sign with it. A negative $b$ gives a positive vertex $x\mathclose{.}$

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Answer: A

Explanation

Find where the minimum happens, then what it is. Those are two different questions and the options here punish confusing them.

$$x = -\frac{b}{2a} = -\frac{6}{2} = -3$$

Now substitute to get the value:

$$y = (-3)^2 + 6(-3) + 1 = 9 - 18 + 1 = -8$$

Check either side: at $x = -4\mathclose{,}$ $y = 16 - 24 + 1 = -7\mathclose{;}$ at $x = -2\mathclose{,}$ $y = 4 - 12 + 1 = -7\mathclose{.}$ Both higher ✓

Why each wrong option is wrong:

• B, $-3$ — the $x$-coordinate of the vertex, which is where the minimum is, not what it is.
• C, $1$ — the constant term, which is $y$ at $x = 0\mathclose{,}$ not at the vertex.
• D, $28$ — substituted $+3$ instead of $-3\mathclose{:}$ $9 + 18 + 1\mathclose{.}$ The vertex is at $-\frac{b}{2a}\mathclose{,}$ and the minus is part of it.

Takeaway: "Where is the minimum" wants $x\mathclose{;}$ "what is the minimum" wants $y\mathclose{.}$ Find $x = -\frac{b}{2a}\mathclose{,}$ then substitute — and read the question again to see which of the two numbers it asked for.

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Section 4 — Nonlinear Systems and Other Equations

A nonlinear system is one straight line and one curve, or two curves. Solve by substitution: rearrange the linear equation to "$y = \ldots$", put it into the other, and you are left with a quadratic in $x\mathclose{.}$

Geometrically you are finding where a line meets a parabola, so expect two intersection points, sometimes one, sometimes none — and the discriminant tells you which before you solve.

Two other equation types appear often enough to be worth their own warning:

• Radical equations. Isolate the root and square both sides. Squaring can create solutions that do not work, so every answer must be checked in the original equation.
• Equations with the variable in a denominator. Multiply through by the denominator, but discard any answer that would make a denominator zero.

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Answer: C

Explanation

Both equations give $y\mathclose{,}$ so set them equal:

$$x^2 - 3 = 2x$$

Move everything to one side:

$$x^2 - 2x - 3 = 0$$

Factor — two numbers multiplying to $-3$ and adding to $-2\mathclose{,}$ so $-3$ and $1\mathclose{:}$

$$(x - 3)(x + 1) = 0 \quad\Longrightarrow\quad x = 3 \text{ or } x = -1$$

Their sum is $3 + (-1) = 2\mathclose{.}$

Or use the shortcut: sum of roots $= -\frac{b}{a} = -\frac{-2}{1} = 2$ ✓

Check one point fully: at $x = 3\mathclose{,}$ the parabola gives $9 - 3 = 6$ and the line gives $6$ ✓ They do meet there.

Why each wrong option is wrong:

• A, $-3$ — read the $-3$ out of the original parabola.
• B, $-2$ — dropped a sign in $-\frac{b}{a}\mathclose{.}$
• D, $3$ — gave one root rather than the sum of both.

Takeaway: Two equations both equal to $y$ — set them equal to each other, collect on one side, and solve the quadratic. Then check whether the question wants the roots, their sum, or the points.

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Answer: D

Explanation

The root is already alone, so square both sides:

$$2x + 7 = (x - 4)^2 = x^2 - 8x + 16$$

Collect:

$$0 = x^2 - 10x + 9 = (x - 1)(x - 9)$$

So the squared equation gives $x = 1$ or $x = 9\mathclose{.}$ Now check both in the original, which is not optional here:

$$x = 9: \quad \sqrt{18 + 7} = \sqrt{25} = 5, \text{ and } 9 - 4 = 5 \quad ✓$$ $$x = 1: \quad \sqrt{2 + 7} = \sqrt{9} = 3, \text{ but } 1 - 4 = -3 \quad \text{no}$$

$x = 1$ fails. The square root symbol means the positive root, so the left side can never be negative — but squaring destroys that information, because $3^2$ and $(-3)^2$ are both 9. A solution created this way is called extraneous, and discarding it is part of the method, not an afterthought.

So the only solution is $x = 9\mathclose{.}$

Why each wrong option is wrong:

• A, $1$ — kept the extraneous root and discarded the real one.
• B, $1$ and $9$ — kept both, skipping the check entirely. This is the option most people pick.
• C, $-1$ and $9$ — a sign slip in the factoring, then kept both.

Takeaway: Squaring both sides can invent solutions. Always substitute back into the original radical equation. A quick filter: if the side without the root comes out negative, that value cannot work.

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Answer: 2

Explanation

Gather the fractions on one side and the numbers on the other:

$$\frac{12}{x} - \frac{8}{x} = 3 - 1 \quad\Longrightarrow\quad \frac{4}{x} = 2$$

Multiply both sides by $x\mathclose{:}$

$$4 = 2x \quad\Longrightarrow\quad x = 2$$

Check in the original: $\frac{12}{2} + 1 = 7$ and $\frac{8}{2} + 3 = 7$ ✓

And check the denominators: $x = 2$ is not zero, so the answer is valid. With a variable on the bottom, an answer of $x = 0$ would have to be thrown away no matter how correctly it was derived.

This is a grid-in. Enter 2.

Takeaway: With the same denominator on both sides, combine the fractions first — it is quicker than clearing them. Then confirm your answer does not make any denominator zero.

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Section 5 — Mixed Practice

Factoring, the formula, the discriminant and the vertex, shuffled.

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Answer: A

Explanation

With no $x$ term, take square roots — and take both:

$$x^2 = 49 \quad\Longrightarrow\quad x = \pm 7$$

Or factor as a difference of squares: $(x+7)(x-7) = 0\mathclose{.}$

B kept only the positive root, which is the mistake this question exists for. C never took the root. D factored as if the constant were attached to an $x\mathclose{.}$

Takeaway: $x^2 = k$ has two answers, $+\sqrt{k}$ and $-\sqrt{k}\mathclose{.}$ The $\pm$ is not decoration.

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Answer: B

Explanation

For $ax^2 + bx + c = 0\mathclose{,}$ the product of the solutions is $\frac{c}{a}\mathclose{:}$

$$\frac{9}{3} = 3$$

Check the long way. Divide through by 3: $x^2 - 4x + 3 = 0\mathclose{,}$ which factors as $(x-1)(x-3)\mathclose{,}$ giving $x = 1$ and $x = 3\mathclose{.}$ Their product is 3 ✓ and their sum is 4, which is option C.

C is the sum, $-\frac{b}{a} = 4\mathclose{.}$ D read $c$ without dividing by $a\mathclose{.}$ A has the sum's sign wrong.

Takeaway: Sum $= -\frac{b}{a}\mathclose{,}$ product $= \frac{c}{a}\mathclose{.}$ Both need the division by $a$ — forgetting it is what makes option D tempting.

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Answer: 8

Explanation

Move everything to one side:

$$x^2 - 5x - 24 = 0$$

Two numbers multiplying to $-24$ and adding to $-5\mathclose{:}$ $-8$ and $3\mathclose{.}$

$$(x - 8)(x + 3) = 0 \quad\Longrightarrow\quad x = 8 \text{ or } x = -3$$

The largest is 8. Check: $64 = 40 + 24$ ✓

This is a grid-in. Enter 8.

Takeaway: Zero on one side before factoring, always. Then read whether the question wants the largest, the smallest, the sum or both.

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Answer: A

Explanation

Compute the discriminant:

$$b^2 - 4ac = 3^2 - 4(2)(5) = 9 - 40 = -31$$

Negative, so there are no real solutions. The parabola sits entirely above the $x$-axis and never crosses it.

B is the zero-discriminant case. C assumes two, which is the usual number but not guaranteed. D cannot happen at all: a quadratic never has more than two solutions.

Takeaway: Compute $b^2 - 4ac$ before solving. Negative means none — and it saves you from grinding through a formula that will produce a root of a negative number.

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Answer: C

Explanation

Two routes, and they agree.

By formula: $x = -\frac{b}{2a} = -\frac{-2}{2} = 1\mathclose{.}$

By symmetry: factor to find the crossings. $x^2 - 2x - 15 = (x-5)(x+3)\mathclose{,}$ so the graph crosses at $x = 5$ and $x = -3\mathclose{.}$ A parabola is symmetric, so the vertex sits exactly halfway between its two $x$-intercepts:

$$\frac{5 + (-3)}{2} = \frac{2}{2} = 1 \; ✓$$

The second route is worth having. When a question gives you the two roots or the two crossings, averaging them is faster than the formula and harder to get wrong.

A read $b$ directly. B dropped the leading minus. D forgot to divide.

Takeaway: The vertex is halfway between the two $x$-intercepts. Average the roots — no formula, no sign errors.

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Answer: D

Explanation

"On the ground" means $h = 0\mathclose{:}$

$$-5t^2 + 20t = 0$$

Factor out the common $-5t\mathclose{:}$

$$-5t(t - 4) = 0 \quad\Longrightarrow\quad t = 0 \text{ or } t = 4$$

Both are real moments: $t = 0$ is when it was thrown, $t = 4$ is when it lands. The question asks when it returns, so $t = 4\mathclose{.}$

Check: at $t = 3\mathclose{,}$ $h = -45 + 60 = 15$ m — still in the air ✓ At $t = 4\mathclose{,}$ $h = -80 + 80 = 0$ ✓

A is the launch. C is the highest point, at $t = -\frac{20}{2(-5)} = 2$ — worth noting that the peak is exactly halfway between the two ground times. B divided the 5 by the 20, where the working divides 20 by 5.

Takeaway: In a height model, $h = 0$ gives two times: the start and the landing. Read which one the question wants, and remember the peak is halfway between them.

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Answer: 0

Explanation

Set the two expressions equal, since both give $y\mathclose{:}$

$$x^2 + 4 = 4x + c$$

Collect everything on one side:

$$x^2 - 4x + (4 - c) = 0$$

"Exactly one point" means this quadratic has exactly one solution, so its discriminant is zero:

$$(-4)^2 - 4(1)(4 - c) = 0$$ $$16 - 16 + 4c = 0 \quad\Longrightarrow\quad 4c = 0 \quad\Longrightarrow\quad c = 0$$

Check: with $c = 0$ the equation is $x^2 - 4x + 4 = (x-2)^2 = 0\mathclose{,}$ whose only solution is $x = 2\mathclose{.}$ At $x = 2$ the parabola gives $8$ and the line gives $8$ — they touch at $(2, 8)$ ✓

A line meeting a parabola at exactly one point is a tangent, and "tangent to" in a question is the same instruction as "exactly one intersection": set the discriminant to zero.

This is a grid-in. Enter 0.

Takeaway: Line and curve intersecting once ⇒ set them equal, collect, and set the discriminant of the result to zero. "Tangent" means the same thing.

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Answer: B

Explanation

The factored form gives the roots immediately: $x = -6$ and $x = 2\mathclose{.}$ The vertex is halfway between them:

$$\frac{-6 + 2}{2} = \frac{-4}{2} = -2$$

Check: at $x = -2\mathclose{,}$ $f = (4)(-4) = -16\mathclose{.}$ At the roots $f = 0\mathclose{,}$ so $-2$ is indeed lower ✓

C and A are the roots themselves — where $f$ is zero, not where it is least. D read a bracket without the sign flip.

Takeaway: Roots from the brackets, vertex halfway between. Where a parabola is zero and where it is least are different questions.

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Answer: D

Explanation

Move the constant across, then complete the square. Halve the middle coefficient and square it: $\left(\frac62\right)^2 = 9\mathclose{.}$

$$x^2 + 6x = 7$$ $$x^2 + 6x + 9 = 7 + 9$$ $$(x + 3)^2 = 16$$

So $a = 3$ and $b = 16\mathclose{.}$

Whatever you add to the left must be added to the right — that is why $b$ is $7 + 9$ and not either number alone.

B forgot to add the 9 to the right as well. C gave the 9 by itself. A gave $a\mathclose{.}$

Takeaway: Halve, square, add to both sides. Then read whether the question wants $a$ or $b\mathclose{.}$

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Answer: A

Explanation

Roots $-4$ and $6$ come from factors $(x + 4)$ and $(x - 6)$ — the signs flip. Multiply out:

$$(x + 4)(x - 6) = x^2 - 6x + 4x - 24 = x^2 - 2x - 24$$

Check both: $16 + 8 - 24 = 0$ ✓ and $36 - 12 - 24 = 0$ ✓

A faster check uses the shortcuts. The sum of the roots is $-4 + 6 = 2\mathclose{,}$ and sum $= -\frac ba\mathclose{,}$ so $b = -2\mathclose{.}$ The product is $-24\mathclose{,}$ and product $= \frac ca\mathclose{,}$ so $c = -24\mathclose{.}$ That gives $x^2 - 2x - 24$ in one line.

B has the middle sign wrong. C used roots 4 and 6. D has the right two numbers in the wrong places: the 2 belongs to the $x$ term and the 24 is the constant.

Takeaway: Root $r$ gives factor $(x - r)$ — the sign flips. Or build it from sum $= -\frac ba$ and product $= \frac ca\mathclose{.}$

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Answer: D

Explanation

The graph crosses the $x$-axis at $x = -1$ and at $x = 5\mathclose{.}$ The parabola opens upward (the $x^2$ coefficient is positive), so it dips below the axis between its two crossings and sits above outside them.

$y < 0$ is the dip:

$$-1 < x < 5$$

The inequality is strict, so the crossings themselves are excluded — at $x = -1$ and $x = 5$ the value is exactly 0, not less than 0.

Test three points: at $x = 0\mathclose{,}$ $y = -5$ ✓ below. At $x = 6\mathclose{,}$ $y = 7$ ✓ above. At $x = -1\mathclose{,}$ $y = 0$ ✓ on the axis.

A includes the two crossings, where $y$ is exactly 0 and so not less than 0. B is the answer to $y > 0\mathclose{.}$ C flipped the signs of the roots.

Takeaway: Upward parabola: negative between the roots, positive outside. Sketch it, or test one point in the middle — that settles which region you want.

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Answer: 16

Explanation

Set them equal:

$$x^2 = 3x + 4 \quad\Longrightarrow\quad x^2 - 3x - 4 = 0 \quad\Longrightarrow\quad (x - 4)(x + 1) = 0$$

So $x = 4$ or $x = -1\mathclose{.}$ A point in the first quadrant has a positive $x\mathclose{,}$ so $x = 4\mathclose{,}$ and then

$$y = 4^2 = 16$$

Check it on the line as well: $3(4) + 4 = 16$ ✓ The other point is $(-1, 1)\mathclose{,}$ which is in the second quadrant.

Stopping at $x = 4$ is the mistake this question is built around. Enter 16.

Takeaway: Two equations equal to $y$ — set them equal and collect on one side. Then read what was asked: an intersection is a point, and the question may want its $y\mathclose{.}$

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Answer: C

Explanation

The greatest height is the vertex, at

$$t = -\frac{b}{2a} = -\frac{19.6}{2(-4.9)} = \frac{19.6}{9.8} = 2 \text{ seconds}$$

Check either side: $h(1) = 14.7\mathclose{,}$ $h(2) = 19.6\mathclose{,}$ $h(3) = 14.7$ ✓ Highest at 2.

B halved once too often, dividing by $2 \times 9.8\mathclose{.}$ A divided the coefficients the wrong way up, $\frac{4.9}{19.6}\mathclose{.}$ D, 4 seconds, is when the ball lands — and notice the peak sits exactly halfway between launch at $t = 0$ and landing at $t = 4\mathclose{,}$ which is a useful check in itself.

Takeaway: Greatest height means the vertex, $t = -\frac{b}{2a}\mathclose{.}$ With a negative $a$ the parabola opens downward, so the vertex is a maximum — and it falls halfway between the two ground times.

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Answer: C

Explanation

Factor out the common $3x\mathclose{:}$

$$3x(x - 4) = 0 \quad\Longrightarrow\quad x = 0 \text{ or } x = 4$$

B is the trap this question exists for. Dividing both sides by $x$ gives $3x - 12 = 0$ and $x = 4$ — but dividing by $x$ assumes $x$ is not zero, and throws away the root $x = 0\mathclose{.}$ Never divide an equation by the variable; factor instead.

A has the sign of the second root wrong: $x - 4 = 0$ gives $+4\mathclose{.}$ D treated it as a difference of two squares, which needs a constant term, not an $x$ term.

Takeaway: Take the common factor out; do not divide by $x\mathclose{.}$ Dividing by something that might be zero silently deletes a solution.

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Answer: B

Explanation

A point on the $x$-axis has $y = 0\mathclose{,}$ so this asks how many real solutions $4x^2 - 12x + 9 = 0$ has. The discriminant:

$$b^2 - 4ac = (-12)^2 - 4(4)(9) = 144 - 144 = 0$$

Zero, so there is exactly one. The quadratic is a perfect square, $(2x - 3)^2\mathclose{,}$ which is zero only at $x = \frac32\mathclose{:}$ the parabola comes down, touches the axis there, and turns back without crossing.

A is the negative-discriminant case; zero is not negative. C assumes every parabola meets the axis twice, which is the usual case and not a guaranteed one. D cannot happen: a quadratic has at most two real roots.

Takeaway: Points shared with the $x$-axis are real roots, so the discriminant counts them: positive two, zero one, negative none.

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