Chapter 3 — Algebra

Answer: D

Explanation

Expand both sides first:

$$10x - 15 = 3x + 12 + 7 = 3x + 19$$

Gather the $x$ terms on the left and the numbers on the right:

$$7x = 34 \qquad x = \frac{34}{7}$$

An awkward fraction is not a sign of an error. Check by substituting: the left gives $5\left(\frac{68}{7} - 3\right) = 5 \times \frac{47}{7} = \frac{235}{7}\mathclose{,}$ and the right gives $\frac{102}{7} + 19 = \frac{235}{7}\mathclose{.}$ Correct.

Why each wrong option is wrong:

• A, $\frac{4}{7}$ — moved the $-15$ across the equals sign without changing its sign, reaching $7x = 19 - 15\mathclose{.}$
• B, $2$ — multiplied only the first term in each bracket, writing $10x - 3 = 3x + 4 + 7\mathclose{.}$
• C, $\frac{20}{7}$ — expanded correctly and then subtracted the 7 where it should be added, reaching $3x + 5$ on the right.

Takeaway: the bracket multiplies everything inside it. Expand fully on both sides before you gather anything.

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Answer: F

Explanation

Three things have to be read off the graph: the line, which side is shaded, and whether the line itself is included.

The line. It crosses the $y$-axis at 2, and from there it falls 1 for every 1 it runs — through $(1, 1)$ and $(2, 0)\mathclose{.}$ So the boundary is $y = -x + 2\mathclose{.}$

The side. Test a point that is clearly in the shading, such as $(3, 3)\mathclose{.}$ On the line, $x = 3$ gives $y = -1\mathclose{;}$ the point has $y = 3\mathclose{,}$ which is greater. The shading is where $y$ is greater than $-x + 2\mathclose{.}$

The line itself. It is drawn solid, so its points are solutions too: the sign is $\ge\mathclose{,}$ not $>\mathclose{.}$

$$y \ge -x + 2$$

Why each wrong option is wrong:

• G, $y \le -x + 2$ — shades the other side. It would put the origin in the region, and the origin is in the white part of the graph.
• H, $y > -x + 2$ — the right side, but a strict sign, which is drawn as a dashed line.
• J, $y \ge x + 2$ — the slope with the wrong sign. The line $y = x + 2$ rises to the right; the one drawn falls, and passes through $(2, 0)\mathclose{.}$

Takeaway: line first, then one test point for the side, then solid or dashed for the sign. Solid means the line is included $(\le$ or $\ge)\mathclose{;}$ dashed means it is not $(<$ or $>)\mathclose{.}$

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Answer: B

Explanation

Both sides are of the form $(\text{number})x + (\text{number})\mathclose{.}$

"No solution" means the $x$ terms cancel — leaving no $x$ to solve for — while the constants disagree. The $x$ terms cancel exactly when their coefficients match:

$$k = 4$$

Then the equation becomes $4x + 9 = 4x - 3\mathclose{.}$ Subtract $4x$ from both sides:

$$9 = -3$$

which is false whatever $x$ is. So there is no solution.

Contrast with infinitely many solutions, which the ACT also asks: that needs the coefficients to match and the constants to match too. Same first condition, opposite conclusion — the constants decide which one you get.

Why each wrong option is wrong:

• A, $9$ and C, $-3$ — matched a constant rather than the coefficient.
• D, $-4$ — right size, wrong sign; $k = -4$ gives $8x = -12\mathclose{,}$ which has one solution.

Takeaway: no solution means the $x$ terms cancel and the numbers do not. Infinitely many means both cancel. Check the constants to tell them apart.

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Answer: J

Explanation

Read the picture first. The heavy line runs from $-3$ to $3\mathclose{.}$ The circle at $-3$ is open, so $-3$ is not a solution; the circle at $3$ is filled, so $3$ is. The number line shows

$$-3 < x \le 3$$

Now solve the options until one gives exactly that. A three-part inequality is solved by doing the same thing to all three parts at once. For J:

$$-5 < 3x + 4 \le 13$$

Subtract 4 from every part: $-9 < 3x \le 9\mathclose{.}$ Divide every part by 3, which is positive, so no sign flips: $-3 < x \le 3$ ✓

Each end keeps whatever sign it started with — the left stayed strict, the right stayed inclusive — and that is what the open and the filled circle record.

Why each wrong option is wrong:

• F, $-5 \le 3x + 4 < 13$ — the right numbers, with the two signs the wrong way round. It gives $-3 \le x < 3\mathclose{:}$ a filled circle at $-3$ and an open one at $3\mathclose{.}$
• G, $-5 < 3x - 4 \le 13$ — the 4 is subtracted, not added, so every part has 4 added back: $-1 < 3x \le 17\mathclose{,}$ and the line would run from $-\tfrac{1}{3}$ to $\tfrac{17}{3}\mathclose{.}$
• H, $-3 < 3x + 4 \le 3$ — copies $-3$ and $3$ off the number line into the inequality. Those are the ends for $x\mathclose{,}$ not for $3x + 4\mathclose{;}$ solving it gives $-\tfrac{7}{3} < x \le -\tfrac{1}{3}\mathclose{.}$

Takeaway: turn the picture into an inequality for $x$ first — open circle strict, filled circle inclusive — and then solve the options until one matches. Do every step to all three parts.

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Section 2 — Systems of Equations

Two equations, two unknowns. Two methods, and the choice between them is worth making deliberately:

Substitution — best when one equation already gives a variable on its own. Put that expression into the other equation.

Elimination — best when neither does. Add or subtract the two equations so one variable disappears, scaling one or both first if needed.

The ACT also asks about systems with no solution (parallel lines: same slope, different intercept) and infinitely many (the same line twice).

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Answer: B

Explanation

Add the two equations. The $+2y$ and $-2y$ cancel:

$$(3x + 5x) + (2y - 2y) = 19 + 13$$

$$8x = 32 \qquad x = 4$$

Check by finding $y$ too: $3(4) + 2y = 19$ gives $2y = 7\mathclose{,}$ so $y = 3.5\mathclose{.}$ Put both into the second equation: $5(4) - 2(3.5) = 20 - 7 = 13\mathclose{.}$ Correct.

Why each wrong option is wrong:

• A, $3.5$ — solved correctly and gave $y\mathclose{.}$ Always check which letter the question asked for; this is the most common way to lose a mark you have earned.
• C, $8$ — added the coefficients of $x$ and stopped.
• D, $32$ — added the two right-hand sides to get 32 and never divided by 8.

Takeaway: when the coefficients of one variable are equal and opposite, add. When they are equal and the same, subtract. Then answer for the letter asked for.

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Answer: F

Explanation

To remove $x\mathclose{,}$ make its coefficients match. Multiply the first equation by 3 and the second by 2:

$$6x + 15y = 72$$ $$6x - 4y = -4$$

Now subtract the second from the first:

$$19y = 76 \qquad y = 4$$

Check: from the first equation $2x + 20 = 24\mathclose{,}$ so $x = 2\mathclose{;}$ and $3(2) - 2(4) = 6 - 8 = -2\mathclose{.}$ Correct.

Why each wrong option is wrong:

• H, $2$ — gave $x\mathclose{,}$ which is 2.
• G, $\frac{68}{19}$ — subtracted the right-hand sides as $72 - 4 = 68\mathclose{.}$ The second one is $-4\mathclose{,}$ and $72 - (-4) = 76\mathclose{.}$
• J, $-2$ — copied the $-2$ from the right of the second equation.

Takeaway: to eliminate a variable, multiply each equation by the other's coefficient. Then subtract if the signs match and add if they do not.

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Answer: D

Explanation

Parallel lines have the same slope. Rearrange each into $y = mx + c$ form.

From the first: $-2y = -6x + 10\mathclose{,}$ so $y = 3x - 5\mathclose{,}$ slope 3.

From the second: $ky = -9x + 3\mathclose{,}$ so $y = -\frac{9}{k}x + \frac{3}{k}\mathclose{,}$ slope $-\frac{9}{k}\mathclose{.}$

Set the slopes equal:

$$-\frac{9}{k} = 3 \qquad \Longrightarrow \qquad k = -3$$

Check that the lines are not identical, which would give infinitely many solutions instead: with $k = -3$ the second equation is $9x - 3y = 3\mathclose{,}$ or $y = 3x - 1\mathclose{.}$ Same slope, different intercept $(-1$ against $-5)\mathclose{,}$ so they are parallel and distinct. No solution. Correct.

Why each wrong option is wrong:

• A, $3$ — got the size right and the sign wrong.
• B, $- \frac{4}{3}$ — divided the wrong pair of coefficients.
• C, $-2$ — took a coefficient from the first equation.

Takeaway: no solution means equal slopes and different intercepts. Always check the intercept too — equal slopes and equal intercepts is the opposite answer, infinitely many.

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Section 3 — Expanding and Factoring

Expanding and factoring are the same operation in opposite directions, and the ACT asks for both.

Three patterns are worth knowing on sight, because spotting them saves the time that the ACT is really testing:

$$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a - b)^2 = a^2 - 2ab + b^2$$ $$(a + b)(a - b) = a^2 - b^2$$

The middle term in the first two is what people forget. $(a+b)^2$ is not $a^2 + b^2\mathclose{.}$

To factor $x^2 + bx + c\mathclose{,}$ find two numbers that multiply to $c$ and add to $b\mathclose{.}$

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Answer: J

Explanation

$(2x - 3)^2$ means $(2x - 3)(2x - 3)\mathclose{.}$ Multiply every term by every term:

$$2x \cdot 2x = 4x^2 \qquad 2x \cdot (-3) = -6x$$ $$-3 \cdot 2x = -6x \qquad -3 \cdot (-3) = +9$$

There are two middle terms, both $-6x\mathclose{,}$ so together they make $-12x\mathclose{:}$

$$(2x - 3)^2 = 4x^2 - 12x + 9$$

Check at $x = 1\mathclose{:}$ the original is $(-1)^2 = 1\mathclose{,}$ and $4 - 12 + 9 = 1\mathclose{.}$ Correct.

Why each wrong option is wrong:

• F, $4 x^{2} + 9$ — squared each term separately. There is always a middle term.
• G, $4 x^{2} - 9$ — used the difference-of-squares pattern, which is for $(a+b)(a-b)\mathclose{,}$ not for a square.
• H, $4 x^{2} - 6 x + 9$ — found one middle term instead of two.

Takeaway: $(a \pm b)^2 = a^2 \pm 2ab + b^2\mathclose{.}$ Square the whole first term, coefficient included, and double the middle.

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Answer: C

Explanation

With a leading coefficient, the two numbers must multiply to $2 \times (-3) = -6$ and add to $+5\mathclose{.}$ That pair is $+6$ and $-1\mathclose{.}$

Split the middle term using them:

$$2x^2 + 6x - x - 3$$

Factor in pairs:

$$2x(x + 3) - 1(x + 3) = (x + 3)(2x - 1)$$

So the factors are $x + 3$ and $2x - 1\mathclose{.}$ Of the options, $2x - 1$ is there.

Check by expanding: $(x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\mathclose{.}$ Correct.

Why each wrong option is wrong:

• A, $x - 3$ — used factors of 3 alone, ignoring the leading 2.
• B, $2 x + 1$ — right numbers with the signs swapped; $(x-3)(2x+1)$ expands to $2x^2 - 5x - 3\mathclose{,}$ with the middle sign wrong.
• D, $x + 1$ — not a factor at all.

Takeaway: with a leading coefficient $a\mathclose{,}$ look for two numbers multiplying to $a \times c$ and adding to $b\mathclose{,}$ then split the middle term and factor in pairs.

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Answer: F

Explanation

$9x^2 - 49$ is a difference of two squares: $9x^2 = (3x)^2$ and $49 = 7^2\mathclose{,}$ with a minus between them and nothing else.

$$a^2 - b^2 = (a - b)(a + b)$$

with $a = 3x$ and $b = 7\mathclose{:}$

$$9x^2 - 49 = (3x - 7)(3x + 7)$$

Check by expanding: $9x^2 + 21x - 21x - 49\mathclose{,}$ and the middle terms cancel — which is exactly why the pattern has no middle term.

Why each wrong option is wrong:

• G, $\left(3 x - 7\right)^{2}$ — $(3x-7)^2$ expands to $9x^2 - 42x + 49\mathclose{,}$ which has a middle term and the wrong constant sign.
• H, $\left(x + 7\right) \left(9 x - 7\right)$ and J, $\left(3 x - 49\right) \left(3 x + 1\right)$ — took a square root of only one part.

Takeaway: two perfect squares with a minus between them and no middle term factors as $(a-b)(a+b)\mathclose{.}$ Take the square root of both parts.

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Section 4 — Quadratic Equations

A quadratic equation is one with an $x^2$ and no higher power. Three ways to solve one, in the order you should try them:

1. Factor, if it factors easily. Fastest by far.
2. Take square roots, if there is no middle term.
3. The quadratic formula, which always works:

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

The ACT does not print that formula. Learn it.

The part under the root, $b^2 - 4ac\mathclose{,}$ is the discriminant, and it tells you how many real solutions there are before you find any of them:

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Answer: B

Explanation

On the ground the height is zero:

$$-t^2 + 5t + 14 = 0$$

Multiply through by $-1$ so that the squared term is positive, then find two numbers multiplying to $-14$ and adding to $-5\mathclose{.}$ They are $-7$ and $+2\mathclose{:}$

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

A product is zero when one of its factors is zero, so $t = 7$ or $t = -2\mathclose{.}$ A time before the throw makes no sense here, so the answer is $t = 7\mathclose{.}$

Why each wrong option is wrong:

• A, $14$ — the height the ball starts from, $h$ at $t = 0\mathclose{.}$ That is a height, not a time.
• C, $2.5$ — the time of the highest point, $-\frac{b}{2a}\mathclose{.}$ The ball is at the top of its flight then, not on the ground.
• D, $-2$ — the negative root. It solves the equation, and the situation rules it out.

Takeaway: factor, set each bracket to zero, then check the context. A length, a time or a count cannot be negative, and the ACT relies on you noticing.

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Answer: G

Explanation

Exactly one real solution means the discriminant is zero:

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

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

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

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

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

Why each wrong option is wrong:

• F, $4$ — took the square root of 16 rather than of $4ac\mathclose{.}$
• H, $16$ — copied the constant.
• J, $64$ — computed $4ac = 64$ and stopped before the square root.

Takeaway: one solution means $b^2 = 4ac\mathclose{.}$ And when a squared quantity is solved, there are two answers — read the question to see which sign it wants.

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Answer: B

Explanation

Take the square root of both sides, remembering both signs:

$$x - 3 = \pm 5$$

That is two separate equations:

$$x - 3 = 5 \Rightarrow x = 8 \qquad x - 3 = -5 \Rightarrow x = -2$$

Check both: $(8-3)^2 = 25$ and $(-2-3)^2 = (-5)^2 = 25\mathclose{.}$ Both correct.

Why each wrong option is wrong:

• A, $-8$ and $2$ — subtracted the 3 rather than adding it when isolating $x\mathclose{.}$
• C, $-5$ and $5$ — stopped at $x - 3 = \pm 5$ and gave those two numbers. The 3 still has to be added to each.
• D, $2$ and $8$ — got $+8$ right and slipped the sign on the other case.

Takeaway: taking a square root gives $\pm\mathclose{.}$ Write out both equations separately — trying to hold both in your head is where the sign errors come from.

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Section 5 — Radical, Rational and Exponential Equations

Three equation types that each carry one specific hazard.

Radical — square both sides to clear the root, then check every answer in the original equation. Squaring can manufacture solutions that were never there.

Rational (a variable in a denominator) — multiply through by the denominator. Any value that would make a denominator zero is excluded.

Exponential — if both sides can be written with the same base, the exponents must be equal.

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Answer: J

Explanation

Square both sides:

$$x + 6 = x^2$$

Bring everything to one side and factor:

$$x^2 - x - 6 = 0 \quad\Longrightarrow\quad (x - 3)(x + 2) = 0$$

So the squared equation gives $x = 3$ and $x = -2\mathclose{.}$ Both must now be tested in the original, because squaring turns a false statement into a true one: $-2 \ne 2\mathclose{,}$ but $(-2)^2 = 2^2\mathclose{.}$

At $x = 3\mathclose{:}$ $\sqrt{9} = 3\mathclose{.}$ ✓

At $x = -2\mathclose{:}$ $\sqrt{4} = 2\mathclose{,}$ and $2 \ne -2\mathclose{.}$ ✗

The square-root symbol means the non-negative root, so it can never equal a negative number. $x = -2$ is extraneous — manufactured by the squaring.

Why each wrong option is wrong:

• F, $-2$ — kept the extraneous root without testing it.
• G, $1$ — added the two roots of the squared equation.
• H, $2$ — factored $x^2 - x - 6$ as $(x - 2)(x + 3)\mathclose{,}$ with the two signs the wrong way round.

Takeaway: after squaring, every solution is only a candidate. Substitute each back into the original equation. A square root is never negative.

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Answer: D

Explanation

Both 9 and 27 are powers of 3:

$$9 = 3^2 \qquad 27 = 3^3$$

Rewrite both sides:

$$(3^2)^x = (3^3)^4$$ $$3^{2x} = 3^{12}$$

With the same base, the exponents must be equal:

$$2x = 12 \qquad x = 6$$

Check: $9^6 = 531441$ and $27^4 = 531441\mathclose{.}$ Correct.

Why each wrong option is wrong:

• A, $\frac{4}{3}$ — divided the exponents without converting to a common base.
• B, $\frac{8}{3}$ — wrote 9 as $3^3$ and 27 as $3^2\mathclose{,}$ the two the wrong way round, reaching $3x = 8\mathclose{.}$
• C, $4$ — copied the 4 from the right-hand side.

Takeaway: rewrite both sides as powers of the same base, then set the exponents equal. Look for the small base hiding inside both numbers — usually 2 or 3.

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Answer: F

Explanation

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

$$\frac{12}{x} - \frac{8}{x} = 3 - 1$$

$$\frac{4}{x} = 2$$

Multiply both sides by $x$ and divide by 2:

$$4 = 2x \qquad x = 2$$

Check in the original: $\frac{12}{2} + 1 = 7$ and $\frac{8}{2} + 3 = 7\mathclose{.}$ Correct. And $x = 2$ does not make any denominator zero, so it is allowed.

Why each wrong option is wrong:

• G, $4$ — subtracted the numerators and stopped, ignoring the constants.
• H, $8$ — reached $\frac{4}{x} = 2$ and multiplied the 4 by 2 instead of dividing.
• J, $20$ — added the numerators.

Takeaway: with the same denominator on both sides, combine the fractions first — it is quicker than multiplying everything out. Then check the answer does not make a denominator zero.

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Section 6 — Rearranging a Formula

"Solve for $R$ in terms of the others" is a standard ACT question and it appeared on both official enhanced forms. There is no new mathematics in it: treat every other letter as if it were a number, and isolate the one you want.

The only real difficulty is that the answer is an expression rather than a number, so there is nothing to substitute back — which makes it worth being slow.

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Answer: B

Explanation

Isolate the term containing $w\mathclose{:}$

$$P - 2l = 2w$$

Divide both sides by 2:

$$w = \frac{P - 2l}{2}$$

Test it with numbers, which is the only real check available. Take $l = 3$ and $w = 4\mathclose{;}$ then $P = 2(3) + 2(4) = 14\mathclose{.}$ Put $P = 14$ and $l = 3$ into the answer: $\frac{14 - 6}{2} = 4\mathclose{.}$ Correct.

Why each wrong option is wrong:

• A, $\frac{P}{2} - 2 l$ — halved $P$ before subtracting, so the $2l$ never got halved.
• C, $\frac{P - l}{2}$ — subtracted $l$ rather than $2l\mathclose{.}$
• D, $P - 2 l$ — subtracted correctly and never divided.

Takeaway: isolate the term with your letter in it, then divide by everything multiplying it. Test with easy numbers — it catches almost every rearranging error in ten seconds.

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Answer: G

Explanation

Multiply both sides by $t$ to bring it up out of the denominator:

$$Ft = mv$$

Now divide by $F\mathclose{:}$

$$t = \frac{mv}{F}$$

Test with numbers: let $m = 2\mathclose{,}$ $v = 3\mathclose{,}$ $t = 6\mathclose{.}$ Then $F = \frac{6}{6} = 1\mathclose{,}$ and the answer gives $t = \frac{2 \times 3}{1} = 6\mathclose{.}$ Correct.

Why each wrong option is wrong:

• F, $F m v$ — multiplied by $F$ instead of dividing.
• H, $\frac{F}{m v}$ — the reciprocal of the right answer, which is what you get by dividing in the wrong direction.
• J, $\frac{F m}{v}$ — moved $m$ but not $v\mathclose{.}$

Takeaway: to free a variable from a denominator, multiply both sides by it first. Then test with three small numbers; an inverted answer shows up immediately.

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Answer: C

Explanation

Square both sides:

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

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

Test both. At $x = 9\mathclose{:}$ $\sqrt{25} = 5$ and $9 - 4 = 5\mathclose{.}$ ✓ At $x = 1\mathclose{:}$ $\sqrt{9} = 3$ but $1 - 4 = -3\mathclose{.}$ ✗

A square root cannot be negative, so $x = 1$ is extraneous.

Why each wrong option is wrong:

• A, $-1$ — factored $x^2 - 10x + 9$ with a sign wrong, as $(x - 9)(x + 1)\mathclose{,}$ and kept the smaller root.
• B, $1$ — kept the extraneous root without testing it.
• D, $10$ — added the two roots of the squared equation.

Takeaway: every solution after squaring is a candidate until substituted back into the original equation.

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Answer: H

Explanation

Cross-multiply, so that each numerator meets the other fraction's denominator:

$$5x = 3(x + 2) = 3x + 6$$

$$2x = 6 \qquad x = 3$$

Check: $\frac{3}{3 + 2} = \frac{3}{5}\mathclose{.}$ ✓ And $x = 3$ does not make the denominator zero, so it is allowed.

Why each wrong option is wrong:

• F, $-5$ — paired each numerator with its own denominator's partner, writing $3x = 5(x + 2)\mathclose{.}$
• G, $1$ — expanded $3(x + 2)$ as $3x + 2\mathclose{,}$ multiplying only the first term.
• J, $6$ — reached $2x = 6$ and stopped.

Takeaway: in a proportion, cross-multiply and expand the bracket fully. Then check the answer does not make a denominator zero.

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Answer: D

Explanation

$$500 \times 1.2 \times 1.2 = 500 \times 1.44 = 720$$

The second hour's growth is calculated on 600, not on 500 — which is why the answer exceeds $500 + 100 + 100\mathclose{.}$

Why each wrong option is wrong:

• A, $600$ — stopped after one hour.
• B, $620$ — grew the wrong base.
• C, $700$ — added 20% of the original twice, which is simple growth, not compound.

Takeaway: compound growth multiplies by $(1 + r)$ once per period. Each period grows from the new amount.

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Answer: G

Explanation

"Decays to 85%" states the multiplier directly: $0.85\mathclose{.}$

$$M = 400(0.85)^h$$

Check one hour: $400 \times 0.85 = 340\mathclose{,}$ a loss of 60, which is 15% of 400.

Why each wrong option is wrong:

• F, $400(0.15)^h$ — used $0.15\mathclose{,}$ which would leave only 15% each hour.
• H, $400 - 0.15h$ — linear decay, losing a fixed 0.15 grams per hour.
• J, $400(1.15)^h$ — a multiplier above 1 grows the sample.

Takeaway: the multiplier is what remains. Decay is below 1, growth above.

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Answer: D

Explanation

Multiply both sides by $(x - 2)\mathclose{:}$

$$6 = 3(x - 2) = 3x - 6$$

$$3x = 12 \qquad x = 4$$

Check: $\frac{6}{4 - 2} = \frac{6}{2} = 3\mathclose{.}$ ✓ And $x = 4$ does not make the denominator zero, so it is allowed.

Why each wrong option is wrong:

• A, $0$ — found $x - 2 = 2$ and then subtracted the 2 instead of adding it.
• B, $2$ — $x = 2$ is precisely the value that is not allowed: it makes the denominator zero and the expression undefined.
• C, $2.5$ — divided the 3 by the 6 instead of the 6 by the 3, reaching $x - 2 = \frac{1}{2}\mathclose{.}$

Takeaway: clear the denominator by multiplying, solve, then confirm the answer does not make any denominator zero.

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Answer: J

Explanation

Find two numbers multiplying to $-40$ and adding to $-3\mathclose{.}$ They are $-8$ and $+5\mathclose{:}$

$$x^2 - 3x - 40 = (x - 8)(x + 5) = 0$$

A product is zero when one factor is zero, so $x = 8$ or $x = -5\mathclose{.}$ The question asks for the positive one, $x = 8\mathclose{.}$

Why each wrong option is wrong:

• F, $-5$ — gave the negative root.
• G, $3$ — read the 3 straight off the equation without factoring.
• H, $5$ — found the right numbers, 5 and 8, but with the signs the wrong way round: $(x - 5)(x + 8)\mathclose{.}$

Takeaway: factor, set each bracket to zero, then read which root the question asks for.

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Answer: C

Explanation

With $a = 3\mathclose{,}$ $b = -5\mathclose{,}$ $c = -2\mathclose{:}$

$$x = \frac{5 \pm \sqrt{25 - 4(3)(-2)}}{6} = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6}$$

$$x = 2 \qquad \text{or} \qquad x = -\tfrac{1}{3}$$

Watch the $-4ac$ term: $c$ is negative, so $-4(3)(-2) = +24\mathclose{,}$ not $-24\mathclose{.}$ That double negative is where the formula usually goes wrong.

Why each wrong option is wrong:

• A, $- \frac{1}{3}$ — the smaller root; the question asked for the larger.
• B, $\frac{5}{6}$ — computed $\frac{5}{6}\mathclose{,}$ forgetting the $\pm\sqrt{\;}$ part entirely.
• D, $4$ — divided $5 + 7$ by $a$ instead of by $2a\mathclose{.}$ The whole bottom of the formula is $2a\mathclose{,}$ here 6.

Takeaway: substitute $a\mathclose{,}$ $b$ and $c$ with their signs, and expect $-4ac$ to be positive whenever $c$ is negative.

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Answer: H

Explanation

Move the constant across and divide by 4:

$$4x^2 = 49 \qquad x^2 = \frac{49}{4}$$

Take the square root of both sides, remembering both signs:

$$x = \pm\frac{7}{2}$$

Check: $4 \times \frac{49}{4} = 49\mathclose{.}$ ✓

Why each wrong option is wrong:

• F, $- \frac{49}{4}$ and $\frac{49}{4}$ — divided by 4 and stopped before the square root.
• G, $-7$ and $7$ — rooted the 49 and ignored the 4.
• J, $- \frac{49}{2}$ and $\frac{49}{2}$ — halved 49 instead of rooting.

Takeaway: no middle term means take square roots. Isolate $x^2$ first, and never lose the $\pm\mathclose{.}$

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Answer: B

Explanation

One squaring is not enough here, because a root sign survives it:

$$x + 5 = \left(\sqrt{x} + 1\right)^2 = x + 2\sqrt{x} + 1$$

The $x$ terms cancel, which leaves the remaining root on its own:

$$4 = 2\sqrt{x} \quad\Longrightarrow\quad \sqrt{x} = 2$$

Square a second time: $x = 4\mathclose{.}$

Check in the original: $\sqrt{4 + 5} = 3$ and $\sqrt{4} + 1 = 3\mathclose{.}$ ✓

Why each wrong option is wrong:

• A, $2$ — found $\sqrt{x} = 2$ and stopped before the second squaring.
• C, $8$ — squared $2\sqrt{x}$ as $2x$ instead of $4x\mathclose{,}$ so $16 = 2x\mathclose{.}$
• D, $16$ — dropped the 2 and read $\sqrt{x} = 4\mathclose{.}$

Takeaway: $\left(\sqrt{x} + 1\right)^2$ is $x + 2\sqrt{x} + 1\mathclose{,}$ not $x + 1\mathclose{.}$ The middle term keeps a root alive, so isolate it and square again, then check the answer in the original equation.

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Answer: G

Explanation

Cross-multiply:

$$3(x + 6) = 12x$$ $$3x + 18 = 12x$$ $$18 = 9x \qquad x = 2$$

Check: $\frac{3}{2} = 1.5$ and $\frac{12}{8} = 1.5\mathclose{.}$ ✓ Neither denominator is zero at $x = 2\mathclose{,}$ so it is allowed.

Why each wrong option is wrong:

• F, $-6$ — $x = -6$ makes the second denominator zero, which is exactly the value the equation excludes.
• H, $6$ — copied the 6 out of the denominator.
• J, $18$ — added the two numbers in the question.

Takeaway: cross-multiply, solve, then confirm the answer does not make any denominator zero.

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