25 hard SAT Nonlinear equations · one variable questions

Real questions from the SAT Climb bank, all at the hard difficulty tier. Pick an answer before you open the explanation. Every question tells you why the answer is right and why each wrong choice is tempting.

Math · Advanced Math~2 per testHard tier

New to this skill? Read the Nonlinear equations · one variable strategy guide first: what it tests, how it is scored, and the traps below in plain English.

What makes these hard

Reports a root when they asked for the sum (or vice versa).
Misses the extraneous-root check.
“Exactly one solution” solved by guessing instead of the discriminant.

Question 1Hard

What is the product of the solutions to $\displaystyle \sqrt{3x + 4} = x$ ?

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Why A is right

Squaring both sides yields $3x + 4$ = $x^{2}$ , which rearranges to $x^{2}$ - $3x - 4 = 0$ . This factors as $(x - 4) (x + 1) = 0$ , giving $x = 4$ or $x = -1$ . Checking $x = 4$ : $\displaystyle \sqrt{3(4} + 4) = \sqrt{16} = 4$ , which equals x, so $x = 4$ is valid. Checking $x = -1$ : $\displaystyle \sqrt{3(-1} + 4) = \sqrt{1} = 1$ , but $x = -1$ , so this is extraneous. Only $x = 4$ is a valid solution, and the product of the valid solutions is 4.

Why the others are wrong

BThis results from a sign error in the constant term when rearranging the squared equation or misapplying Vieta's formulas.
CThis is the product of both algebraic solutions from the factored form (4 × (-1) with a sign error or 4 × 4 from repeating the valid root), including computation errors.
DThis results from finding only one of the algebraic roots or miscounting which solutions are valid after checking.

Question 2Hard

$x^{2} + 6x + k = 0$ . How many distinct real solutions does the given equation have if $k = 9$ ?

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Why B is right

For a quadratic equation $ax^{2}$ + bx + $c = 0$ , the discriminant is $b^{2}$ - 4ac. Here $a = 1$ , $b = 6$ , $c = 9$ , so the discriminant is $6^{2} - 4$ (1)(9) = 36 - 36 = 0. When the discriminant equals zero, there is exactly one distinct real solution.

Why the others are wrong

AThis results from an arithmetic error in computing the discriminant, such as calculating 6² - 4(1)(9) = 36 - 36 as negative instead of zero.
CThis results from incorrectly concluding that the equation has two solutions without computing the discriminant, or from a sign error that makes the discriminant appear positive.
DThis results from treating the equation as an identity rather than recognizing it as a quadratic equation with a parameter.

Question 3Hard

What is the positive solution to $\displaystyle 2x + 3 = \sqrt{7x + 18}$ ?

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Why C is right

Squaring both sides yields $(2x + 3)^{2} = 7x + 18$ , which expands to $4x^{2} + 12x + 9 = 7x + 18$ . Rearranging gives $4x^{2} + 5x$ − 9 = 0. Factoring yields $(4x + 9)$ (x − 1) = 0, so x = −9/4 or $x = 1$ . Checking $x = 1$ : 2(1) + 3 = 5 and $\displaystyle \sqrt{25} = 5$ , which is true. Checking x = −9/4: 2(−9/4) + 3 = −3/2 (negative), but $\displaystyle \sqrt{7(-9/4} + 18) = \sqrt{9/4} = 3/2$ (positive), so this is extraneous. The positive solution is $x = 1$ .

Why the others are wrong

AThis results from incorrectly solving the quadratic or including an extraneous root without verification.
BThis is a sign error from mishandling the algebraic manipulation or incorrectly identifying which root is positive.
DThis omits proper checking and selects an incorrect root from the factored form.

Question 4Hard

$x^{2}$ − kx + 36 = 0. In the given equation, k is a positive constant. If the equation has exactly one solution, what is the value of k?

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Why C is right

A quadratic equation $ax^{2}$ + bx + $c = 0$ has exactly one solution when its discriminant $b^{2}$ − 4ac equals zero. Here, $a = 1$ , b = −k, and $c = 36$ , so (−k) $^{2}$ − 4(1)(36) = 0. This gives $k^{2}$ − 144 = 0, so $k^{2}$ = 144 and $k = 12$ (since k is positive).

Why the others are wrong

AThis results from incorrectly setting k = √36 = 6 instead of solving the discriminant equation k² = 144.
BThis is obtained by incorrectly using k² = 81 (perhaps from a miscalculation of 4 · 36) or confusing the discriminant condition.
DThis results from doubling one of the incorrect values or using k = 36/2 instead of solving k² = 144.

Question 5Hard

What is the product of the solutions to $\displaystyle \sqrt{3x + 7} = x - 1$ ?

Show the answer and explanation

Why C is right

Squaring both sides gives $3x + 7$ = $x^{2}$ - $2x + 1$ , which rearranges to $x^{2}$ - $5x - 6 = 0$ . This factors as $(x - 6) (x + 1) = 0$ , giving $x = 6$ or $x = -1$ . Checking $x = 6$ : $\displaystyle \sqrt{25} = 5$ and 6 - 1 = 5 (valid). Checking $x = -1$ : $\displaystyle \sqrt{4} = 2$ but -1 - 1 = -2 (invalid). Only $x = 6$ is valid, but the product of the algebraic solutions is 6 × (-1) = -6. Since only one solution is valid, the product of valid solutions is 6.

Why the others are wrong

AThis is the product of both algebraic solutions including the extraneous root, which should be excluded.
BThis results from a sign error when computing the product of solutions.
DThis results from using only part of the factorization or missing one of the algebraic solutions.

Question 6Hard

$x^{3} - 7x^{2} + 7x + 15 = 0$ . What is the product of the solutions to the given equation?

Show the answer and explanation

Why C is right

By Vieta's formulas for $x^{3}$ + $bx^{2}$ + cx + $d = 0$ , the product of roots equals -d. For $x^{3}$ - $7x^{2} + 7x + 15 = 0$ , we have $d = 15$ , so the product is -15. We can verify by factoring: testing $x = -1$ gives -1 - 7 - 7 + 15 = 0, so $(x + 1)$ is a factor. Dividing gives $(x + 1) ([MATH]x^{2} - 8x + 15)$ = 0[/MATH], which factors as $(x + 1) (x - 3) (x - 5) = 0$ . The roots are -1, 3, and 5, with product (-1)(3)(5) = -15.

Why the others are wrong

AThis is the coefficient of x, mistakenly used as the product of roots, or the product of only two roots.
BThis results from using the constant term directly without the required sign change in Vieta's formulas.
DThis results from using the coefficient of x with a sign change, or from miscalculating the product of roots.

Question 7Hard

What is the product of the solutions to $x^{3} - 7x^{2} + 7x + 15 = 0$ ?

Show the answer and explanation

Why C is right

By Vieta's formulas for a cubic equation $x^{3}$ + $ax^{2}$ + bx + $c = 0$ , the product of the roots is -c. Here $c = 15$ , so the product is -15. This can be verified by factoring: testing $x = -1$ gives $(-1)^{3} - 7 (-1)^{2} + 7$ (-1) + 15 = -1 - 7 - 7 + 15 = 0, so $x + 1$ is a factor. Dividing gives $(x + 1) ([MATH]x^{2} - 8x + 15)$ = (x + 1) (x - 3) (x - 5) = 0[/MATH], yielding $x = -1$ , $x = 3$ , and $x = 5$ . The product is (-1) × 3 × 5 = -15.

Why the others are wrong

AThis results from a sign error when applying Vieta's formula, taking the product as positive instead of negative.
BThis is the product of only two of the three roots, omitting one factor from the calculation.
DThis incorrectly includes an extraneous factor or miscalculates the product by adding an extra root.

Question 8Hard

$x^{4} - 10x^{2} + 9 = 0$ How many distinct real solutions does the given equation have?

Show the answer and explanation

Why C is right

Let u = $x^{2}$ , giving $u^{2}$ - $10u + 9 = 0$ , which factors as $(u - 9) (u - 1) = 0$ . Thus $u = 9$ or $u = 1$ , so $x^{2}$ = 9 or $x^{2}$ = 1, yielding $x = \pm 3$ or $x = \pm 1$ . There are 4 distinct real solutions: 3, -3, 1, -1.

Why the others are wrong

AThis incorrectly assumes the equation has no real solutions, perhaps from misapplying the discriminant test.
BThis results from only counting the positive roots or missing the fact that each value of u yields two values of x.
DThis results from a fundamental misunderstanding of how quartic equations factor.

Question 9Hard

What is the positive solution to $\displaystyle x + \sqrt{2x - 1} = 8$ ?

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Why B is right

Isolating the radical gives $\displaystyle \sqrt{2x - 1} = 8 - x$ . Squaring both sides yields $2x - 1 = 64 - 16x$ + $x^{2}$ , which rearranges to $x^{2}$ - $18x + 65 = 0$ . This factors as $(x - 13) (x - 5) = 0$ , giving $x = 13$ or $x = 5$ . Checking $x = 13$ : $\displaystyle 13 + \sqrt{25} = 13 + 5 = 18 \ne 8$ (extraneous). Checking $x = 5$ : $\displaystyle 5 + \sqrt{9} = 5 + 3 = 8$ (valid). The positive solution is 5.

Why the others are wrong

AThis results from an error in solving the quadratic equation or incomplete factorization.
CThis value appears during algebraic manipulation but is extraneous when checked in the original equation.
DThis results from a sign error when solving the quadratic.

Question 10Hard

$9x^{2} + kx + 25 = 0$ . What value of k makes the equation have exactly one solution?

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Why B is right

For exactly one solution, the discriminant must be zero. For $9x^{2}$ + kx + 25 = 0, the discriminant is $k^{2}$ – 4(9)(25) = $k^{2}$ – 900. Setting $k^{2}$ – 900 = 0 gives $k^{2}$ = 900, so $k = 30$ or k = –30. The positive value is 30.

Why the others are wrong

AThis results from multiplying the square roots of the coefficients (√9 · √25 = 3 · 5 = 15) instead of using the discriminant condition.
CThis results from incorrectly computing 4(9)(25) as 225 and then setting k = 2√225 = 2(15) = 30, then doubling to 45 through an arithmetic error.
DThis results from setting k equal to 4(9)(25) = 900 and then computing √900 = 30 but incorrectly recording the squared value 225.

Question 11Hard

If $\displaystyle \sqrt{2x + 5} = x - 5$ , what is the value of x?

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Why C is right

Squaring both sides yields $2x + 5 = (x - 5)^{2}$ , which expands to $2x + 5$ = $x^{2}$ - $10x + 25$ . Rearranging gives $x^{2}$ - $12x + 20 = 0$ , which factors as $(x - 2) (x - 10) = 0$ . The solutions are $x = 2$ and $x = 10$ . Checking $x = 2$ : $\displaystyle \sqrt{9} = 3$ but 2 - 5 = -3, so $x = 2$ is extraneous. Checking $x = 10$ : $\displaystyle \sqrt{25} = 5$ and 10 - 5 = 5, so $x = 10$ is valid.

Why the others are wrong

AThis is the extraneous root obtained from squaring but not verified in the original equation, where the square root equals 3 but x - 5 equals -3.
BThis results from setting x - 5 = 0 without solving the full equation after squaring.
DThis comes from an arithmetic error in factoring or solving the quadratic equation.

Question 12Hard

$\displaystyle \sqrt{2x + 9} + 3 = x$ . What is the solution to the given equation?

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Why C is right

Subtracting 3 from both sides gives $\displaystyle \sqrt{2x + 9} = x - 3$ . Squaring both sides gives $2x + 9$ = $x^{2}$ - $6x + 9$ , which simplifies to $x^{2}$ - $8x = 0$ . Factoring yields $x(x - 8) = 0$ , so $x = 0$ or $x = 8$ . Checking $x = 0$ : $\displaystyle \sqrt{9} + 3 = 6$ , but $x = 0$ , so this is extraneous. Checking $x = 8$ : $\displaystyle \sqrt{16 + 9} + 3 = \sqrt{25} + 3 = 5 + 3 = 8$ , which equals x, so $x = 8$ is valid.

Why the others are wrong

AThis is an extraneous root from the squared equation that doesn't satisfy the original equation when checked.
BThis results from a computational error in solving the quadratic or from missing the correct root.
DThis results from a sign error when setting up or solving the quadratic equation.

Question 13Hard

What is the product of the solutions to $x^{3} + 4x^{2} - 5x - 20 = 0$ ?

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Why B is right

Factoring by grouping: $x^{2} (x + 4) - 5 (x + 4) = 0$ , which factors as ( $x^{2}$ - 5) $(x + 4) = 0$ . This gives $\displaystyle (x - \sqrt{5}) (x + \sqrt{5}) (x + 4) = 0$ , so the solutions are $\displaystyle x = \sqrt{5}$ $\displaystyle x = -\sqrt{5}$ and $x = -4$ . The product of all solutions is $\displaystyle (\sqrt{5}) (-\sqrt{5})$ (-4) = (-5)(-4) = 20. By Vieta's formulas, the product of roots equals $-d/a$ where d is the constant term and a is the leading coefficient, giving -(-20)/1 = 20.

Why the others are wrong

AThis incorrectly applies Vieta's formulas by using the constant term directly without accounting for the sign and leading coefficient.
CThis uses the coefficient of x² instead of correctly applying Vieta's formulas for the product of roots.
DThis is the product of only two of the three roots, missing one solution in the calculation.

Question 14Hard

$x^{4} - 13x^{2} + 36 = 0$ How many distinct real solutions does the given equation have?

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Why C is right

Let u = $x^{2}$ . Then $u^{2}$ - $13u + 36 = 0$ , which factors as $(u - 4) (u - 9) = 0$ . So $u = 4$ or $u = 9$ . For $u = 4$ , $x = \pm 2$ . For $u = 9$ , $x = \pm 3$ . All four values are real and distinct.

Why the others are wrong

AThis results from finding only the positive solutions or only considering one value of u.
BThis results from incorrectly including an extraneous solution or miscounting the valid roots.
DThis results from a sign error when evaluating the discriminant or factoring the quadratic in u.

Question 15Hard

What is the positive solution to $\displaystyle x + 4 = \sqrt{8x + 16}$ ?

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Why A is right

Squaring both sides yields $(x + 4)^{2} = 8x + 16$ , which expands to $x^{2}$ + $8x + 16 = 8x + 16$ . Simplifying gives $x^{2}$ = 0, so $x = 0$ . Checking: 0 + 4 = 4 and $\displaystyle \sqrt{16} = 4$ , which is true. The positive solution (or non-negative) is $x = 0$ .

Why the others are wrong

BThis incorrectly identifies an additional root that doesn't satisfy the original equation.
CThis results from algebraic errors when squaring or expanding, leading to an extraneous solution.
DThis is a sign error, taking the negative of a root that should be non-negative.

Question 16Hard

What is the product of the solutions to $x^{2} - 7x + k = 0$ , where k is a constant and the equation has exactly one solution?

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Why B is right

For a quadratic equation to have exactly one solution, the discriminant must equal zero. For $x^{2}$ - $7x + k = 0$ , the discriminant is $49 - 4k = 0$ , so $k = 49/4$ . The product of solutions by Vieta's formulas is $c/a = k/1 = 49/4$ .

Why the others are wrong

AThis incorrectly uses the coefficient of x rather than applying Vieta's formulas for the product.
CThis assumes one root is zero, which would require k = 0, but then the discriminant would be 49, giving two distinct solutions.
DThis results from an arithmetic error in calculating k from the discriminant condition, using 2 instead of 4 in the denominator.

Question 17Hard

$\displaystyle \sqrt{3x + 7} = x - 1$ . What is the product of all solutions to the given equation?

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Why D is right

Squaring both sides gives $3x + 7$ = $x^{2}$ - $2x + 1$ , which simplifies to $x^{2}$ - $5x - 6 = 0$ . Factoring yields $(x - 6) (x + 1) = 0$ , so $x = 6$ or $x = -1$ . Checking $x = -1$ : $\displaystyle \sqrt{3(-1} + 7) = 2$ but -1 - 1 = -2, so $x = -1$ is extraneous. Only $x = 6$ is valid, and the product of all solutions is 6.

Why the others are wrong

AThis is the product of both algebraic solutions (-1)(6) = -6 divided by 3, or computing the product incorrectly with a sign error.
BThis results from taking the product of -1 and 6 as -6, then applying a sign error to get 2, or dividing by an incorrect factor.
CThis is the value obtained by missing the root x = 6 and only considering partial factorization or an incomplete solution set.

Question 18Hard

$x^{3} + 3x^{2} - 4x - 12 = 0$ . What is the product of the solutions to the given equation?

Show the answer and explanation

Why B is right

By Vieta's formulas for a cubic equation $x^{3}$ + $bx^{2}$ + cx + $d = 0$ , the product of roots equals -d. For $x^{3}$ + $3x^{2} - 4x - 12 = 0$ , we have $d = -12$ , so the product is -(-12) = 12. We can verify by factoring: grouping gives $x^{2} (x + 3) - 4 (x + 3) = 0$ , so ( $x^{2}$ - 4) $(x + 3) = 0$ , yielding $x = -2$ , 2, or -3. The product is (-2)(2)(-3) = 12.

Why the others are wrong

AThis results from incorrectly applying Vieta's formulas without the sign change for the constant term.
CThis is the product of only two of the three roots, missing one root in the calculation.
DThis results from taking the absolute value of the coefficient of x or from a sign error in applying Vieta's formulas.

Question 19Hard

What is the sum of the solutions to $\displaystyle \sqrt{x + 12} = x$ ?

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Why D is right

Squaring both sides gives $x + 12$ = $x^{2}$ , which simplifies to $x^{2}$ - $x - 12 = 0$ . Factoring yields $(x - 4) (x + 3) = 0$ , so $x = 4$ or $x = -3$ . Checking $x = -3$ : $\displaystyle \sqrt{-3 + 12} = \sqrt{9} = 3$ , but $-3 \ne 3$ , so this is extraneous. Checking $x = 4$ : $\displaystyle \sqrt{4 + 12} = \sqrt{16} = 4$ ✓. The only valid solution is $x = 4$ , so the sum is 4.

Why the others are wrong

AThis results from taking only part of a solution value or miscalculating after factoring.
BThis is the extraneous solution that satisfies the squared equation but not the original radical equation.
CThis reflects a sign error in computing the sum or applying Vieta's formula incorrectly.

Question 20Hard

$3x^{2} - 18x + n = 0$ . How many distinct real solutions does the given equation have if $n = 27$ ?

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Why B is right

The discriminant is $b^{2}$ - 4ac where $a = 3$ , $b = -18$ , $c = 27$ . Computing: $(-18)^{2} - 4$ (3)(27) = 324 - 324 = 0. Since the discriminant equals zero, there is exactly one distinct real solution.

Why the others are wrong

AThis results from a computational error that makes the discriminant appear negative, such as miscalculating 4(3)(27) as a larger value.
CThis results from failing to compute the discriminant accurately and assuming it is positive, leading to the conclusion of two solutions.
DThis results from incorrectly treating the parameter condition as making every value of x a solution.

Question 21Hard

$4x^{2}$ – $12x + c = 0$ . What value of c makes the equation have exactly one solution?

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Why C is right

For exactly one solution, the discriminant must be zero. For $4x^{2}$ – $12x + c = 0$ , the discriminant is (–12) $^{2}$ – 4(4) $(c) = 144$ – 16c. Setting 144 – $16c = 0$ gives $16c = 144$ , so $c = 9$ .

Why the others are wrong

AThis results from dividing the coefficient –12 by the leading coefficient 4, which does not relate to the discriminant condition.
BThis results from incorrectly simplifying 144 – 16c = 0 as c = 12/2 = 6, failing to correctly isolate c.
DThis results from setting (–12)² = 16c and solving c = 144/4 = 36, omitting the 4(4) factor in the discriminant formula.

Question 22Hard

What is the solution to the equation $\displaystyle \sqrt{4x - 3} + 2 = x$ ?

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Why C is right

Isolating the radical gives $\displaystyle \sqrt{4x - 3} = x - 2$ . Squaring both sides yields $4x - 3$ = $x^{2}$ - $4x + 4$ , which simplifies to $x^{2}$ - $8x + 7 = 0$ . Factoring gives $(x - 7) (x - 1) = 0$ , so $x = 7$ or $x = 1$ . Checking $x = 1$ : $\displaystyle \sqrt{4(1} - 3) + 2 = 1 + 2 = 3 \ne 1$ , so $x = 1$ is extraneous. Checking $x = 7$ : $\displaystyle \sqrt{4(7} - 3) + 2 = \sqrt{25} + 2 = 7$ ✓.

Why the others are wrong

AThis is the extraneous solution that does not satisfy the original equation.
BThis results from a sign error in solving the quadratic equation.
DThis results from missing the correct solution and selecting an incorrect value.

Question 23Hard

$x^{3} - 2x^{2} - 9x + 18 = 0$ . What is the sum of the solutions to the given equation?

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Why C is right

By Vieta's formulas for a cubic equation $x^{3}$ + $bx^{2}$ + cx + $d = 0$ , the sum of roots equals -b. For $x^{3}$ - $2x^{2} - 9x + 18 = 0$ , we have $b = -2$ , so the sum of solutions is -(-2) = 2. We can verify by factoring: grouping gives $x^{2} (x - 2) - 9 (x - 2) = 0$ , so ( $x^{2}$ - 9) $(x - 2) = 0$ , yielding $x = -3$ , 3, or 2. The sum is -3 + 3 + 2 = 2.

Why the others are wrong

AThis results from incorrectly applying Vieta's formulas with the wrong sign for the coefficient of x².
BThis results from missing one of the three roots when factoring and summing only two roots.
DThis is the absolute value of the coefficient of x, mistakenly used as the sum of roots.

Question 24Hard

What is the product of the solutions to $2x^{3} - x^{2} - 18x + 9 = 0$ ?

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Why C is right

For the cubic equation $2x^{3}$ - $x^{2}$ - $18x + 9 = 0$ , by Vieta's formulas the product of the roots is $-d/a = -9/2 = -4.5$ . This can be verified by factoring: $x^{2} (2x - 1) - 9 (2x - 1) = 0$ gives ( $x^{2}$ - 9) $(2x - 1) = 0$ , so $(x - 3) (x + 3) (2x - 1) = 0$ , yielding $x = 3$ , $x = -3$ , and $x = 1/2$ . The product is 3 × (-3) × (1/2) = -9/2 = -4.5.

Why the others are wrong

AThis is the product of only two of the three roots, omitting the third factor from the calculation.
BThis results from a sign error when applying Vieta's formula, incorrectly treating the constant term's sign.
DThis incorrectly multiplies by an extraneous factor, as if the equation had four roots instead of three.

Question 25Hard

$5x^{2} + 20x + q = 0$ . How many distinct real solutions does the given equation have if $q = 15$ ?

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Why C is right

The discriminant is $b^{2}$ - 4ac where $a = 5$ , $b = 20$ , $c = 15$ . Computing: $20^{2} - 4$ (5)(15) = 400 - 300 = 100. Since 100 > 0, there are exactly two distinct real solutions.

Why the others are wrong

AThis results from incorrectly computing 4(5)(15) as a value larger than 400, making the discriminant appear negative.
BThis results from an arithmetic error that makes the discriminant appear to equal zero, such as miscalculating 400 - 300 as 0.
DThis results from misunderstanding the parameter and incorrectly concluding the equation is an identity with infinitely many solutions.

These 25 are a sample, not a study plan

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