20 easy SAT Nonlinear equations · one variable questions

These are the questions most test-takers get right. They are worth practising anyway: on the digital SAT the easy questions in Module 1 are what route you into the harder, higher-scoring Module 2, so dropping one costs more than it looks.

Every question below is a real item from the SAT Climb bank, tagged easy by the same difficulty model the app uses to build your practice. Pick an answer before you open the explanation.

Math · Advanced Math~2 per testEasy tier

None of these repeat the 25 hard Nonlinear equations · one variable questions, so the two pages together are 45 different questions on this one skill.

Question 1Easy

What is the solution to the equation (x+3)(x−7)=0(x + 3) (x - 7) = 0?

Show the answer and explanation

Why B is right

By the zero product property, if (x+3)(x−7)=0(x + 3) (x - 7) = 0, then x+3=0x + 3 = 0 or x−7=0x - 7 = 0. Solving the first equation yields x=−3x = -3, and solving the second equation yields x=7x = 7.

Why the others are wrong

  • AThis results from applying incorrect signs when solving each factor, reversing the signs of both solutions.
  • CThis represents finding only one of the two solutions and missing the second root from the factor (x - 7).
  • DThis results from adding 7 and -3 instead of recognizing that each factor separately equals zero.
Question 2Easy
3x2+8x−3=03x^{2} + 8x - 3 = 0

How many distinct real solutions does the given equation have?

Show the answer and explanation

Why C is right

The discriminant is b2b^{2} - 4ac=64−44ac = 64 - 4(3)(-3) = 64 + 36 = 100. When the discriminant is positive, the quadratic has exactly two distinct real solutions.

Why the others are wrong

  • AThis results from a sign error when computing 4(3)(-3), treating it as positive and getting 64 - 36 = 28 but then incorrectly concluding negative.
  • BThis assumes the discriminant equals zero, perhaps by computing 64 + 36 incorrectly.
  • DThis incorrectly interprets the equation as an identity with infinitely many solutions rather than a quadratic with a finite solution set.
Question 3Easy

What is the positive solution to the equation x2−64=0x^{2} - 64 = 0?

Show the answer and explanation

Why B is right

The equation x2x^{2} - 64 = 0 can be rewritten as x2x^{2} = 64. Taking the square root of both sides yields x=8x = 8 or x=−8x = -8. Since the question asks for the positive solution, the answer is 8.

Why the others are wrong

  • AThis results from taking the square root of 64 incorrectly or confusing the relationship between squared and square root operations.
  • CThis is the negative solution to the equation, but the question specifically asks for the positive solution.
  • DThis results from halving 64 instead of taking its square root, representing a misunderstanding of the inverse operation needed.
Question 4Easy

What is the sum of the solutions to the equation (x−4)(x+7)=0(x - 4) (x + 7) = 0?

Show the answer and explanation

Why B is right

The given equation (x−4)(x+7)=0(x - 4) (x + 7) = 0 has solutions where either factor equals zero. Setting x−4=0x - 4 = 0 gives x=4x = 4, and setting x+7=0x + 7 = 0 gives x=−7x = -7. The sum of the solutions is 4 + (-7) = -3.

Why the others are wrong

  • AThis is the product of the solutions (4 × -7 = -28), not their sum.
  • CThis results from a sign error, finding the sum as 4 - 7 but then taking the absolute value, or incorrectly computing 7 - 4.
  • DThis results from adding the absolute values of the two solutions (4 + 7 = 11) rather than correctly handling the negative sign in -7.
Question 5Easy

What is the sum of the solutions to x2−5x−14=0x^{2} - 5x - 14 = 0?

Show the answer and explanation

Why C is right

The equation x2x^{2} - 5x−14=05x - 14 = 0 factors as (x−7)(x+2)=0(x - 7) (x + 2) = 0, giving solutions x=7x = 7 and x=−2x = -2. The sum of these solutions is 7 + (-2) = 5.

Why the others are wrong

  • AThis results from taking the negative of the coefficient of x instead of recognizing that the sum of roots equals the positive coefficient.
  • BThis is only one of the two solutions, not the sum of both solutions.
  • DThis is the constant term, not the sum of the solutions.
Question 6Easy

What is the positive solution to the equation (x−4)2=49(x - 4)^{2} = 49?

Show the answer and explanation

Why D is right

Taking the square root of both sides yields x−4=7x - 4 = 7 or x−4=−7x - 4 = -7. Solving the first equation gives x=11x = 11, and solving the second gives x=−3x = -3. Since the question asks for the positive solution, the answer is 11.

Why the others are wrong

  • AThis results from taking the square root of 49 without adding back the 4 that was subtracted in the equation.
  • BThis is the negative solution to the equation, but the question specifically asks for the positive solution.
  • CThis results from ignoring the squared term entirely and treating the left side as simply x.
Question 7Easy

What is the product of the solutions to the equation x2+4x−21=0x^{2} + 4x - 21 = 0?

Show the answer and explanation

Why A is right

Factoring yields (x+7)(x−3)=0(x + 7) (x - 3) = 0, so x=−7x = -7 or x=3x = 3. The product of these solutions is (-7)(3) = -21. Alternatively, for a quadratic ax2ax^{2} + bx + c=0c = 0, the product of solutions equals c/ac/a, which is -21/1 = -21.

Why the others are wrong

  • BThis is the sum of the solutions, not the product.
  • CThis results from a sign error when calculating the sum of the solutions and confusing it with the product.
  • DThis results from a sign error when calculating the product of -7 and 3.
Question 8Easy

What is a solution to x2−2x−15=0x^{2} - 2x - 15 = 0?

Show the answer and explanation

Why D is right

The equation x2x^{2} - 2x−15=02x - 15 = 0 factors as (x−5)(x+3)=0(x - 5) (x + 3) = 0, giving solutions x=5x = 5 and x=−3x = -3. Of these, 5 is given as a choice.

Why the others are wrong

  • AThis is the constant term, not a solution to the equation.
  • BThis is the other solution to the equation but presented with incorrect sign context in the distractor set.
  • CThis results from a sign error when solving (x + 3) = 0, incorrectly yielding x = 3 instead of x = -3.
Question 9Easy

What is the positive solution to the equation x2−9=0x^{2} - 9 = 0?

Show the answer and explanation

Why B is right

Adding 9 to both sides of the given equation yields x2x^{2} = 9. Taking the square root of both sides yields x=3x = 3 or x=−3x = -3. Of these two solutions, only 3 is positive.

Why the others are wrong

  • AThis is the negative solution to the equation. The question asks for the positive solution.
  • CThis is the result of not taking the square root after solving for x².
  • DThis incorrectly treats the equation as if it has only one solution at the y-intercept.
Question 10Easy

What is the positive solution to the equation (x−6)(x+2)=0(x - 6) (x + 2) = 0?

Show the answer and explanation

Why D is right

Setting each factor equal to zero gives x−6=0x - 6 = 0 or x+2=0x + 2 = 0, so x=6x = 6 or x=−2x = -2. Since the question asks for the positive solution, the answer is 6.

Why the others are wrong

  • AThis is the product of the solutions (6)(-2), not the positive solution.
  • BThis is the negative solution to the equation, but the question asks for the positive solution.
  • CThis results from confusing the solutions with the sum of the solutions (6 + (-2) = 4).
Question 11Easy

What is a solution to x2+8x+12=0x^{2} + 8x + 12 = 0?

Show the answer and explanation

Why B is right

The equation x2x^{2} + 8x+12=08x + 12 = 0 factors as (x+6)(x+2)=0(x + 6) (x + 2) = 0, giving solutions x=−6x = -6 and x=−2x = -2. Of these, -6 is given as a choice.

Why the others are wrong

  • AThis results from using the constant term directly rather than solving the equation.
  • CThis results from a sign error when solving (x + 2) = 0, yielding x = 2 instead of x = -2.
  • DThis results from confusing the constant term with a solution to the equation.
Question 12Easy

What is the positive solution to the equation x2=121x^{2} = 121?

Show the answer and explanation

Why A is right

Taking the square root of both sides of x2x^{2} = 121 yields x=11x = 11 or x=−11x = -11. Since the question asks for the positive solution, the answer is 11.

Why the others are wrong

  • BThis is the negative solution to the equation, but the question specifically asks for the positive solution.
  • CThis results from dividing 121 by 2 instead of taking its square root.
  • DThis results from doubling 11 instead of recognizing 11 as the direct square root of 121.
Question 13Easy

What is the positive solution to the equation x2=49x^{2} = 49?

Show the answer and explanation

Why B is right

Taking the square root of both sides of x2x^{2} = 49 yields x=7x = 7 or x=−7x = -7. Since the question asks for the positive solution, the answer is 7.

Why the others are wrong

  • AThis is the negative solution to the equation, but the question specifically asks for the positive solution.
  • CThis results from incorrectly multiplying 7 by 2 instead of taking the square root.
  • DThis results from incorrectly dividing 49 by 2 instead of taking the square root.
Question 14Easy

What is the product of the solutions to x2−9x+20=0x^{2} - 9x + 20 = 0?

Show the answer and explanation

Why D is right

The equation x2x^{2} - 9x+20=09x + 20 = 0 factors as (x−5)(x−4)=0(x - 5) (x - 4) = 0, giving solutions x=5x = 5 and x=4x = 4. The product of these solutions is 5 × 4 = 20.

Why the others are wrong

  • AThis results from incorrectly applying a negative sign to the constant term.
  • BThis is only one of the two solutions, not the product of both solutions.
  • CThis is the absolute value of the coefficient of x, not the product of the solutions.
Question 15Easy

What is the positive solution to x2−6x−16=0x^{2} - 6x - 16 = 0?

Show the answer and explanation

Why D is right

The equation x2x^{2} - 6x−16=06x - 16 = 0 factors as (x−8)(x+2)=0(x - 8) (x + 2) = 0, giving solutions x=8x = 8 and x=−2x = -2. Since the question asks for the positive solution, the answer is 8.

Why the others are wrong

  • AThis is the negative solution, but the question specifically asks for the positive solution.
  • BThis results from incorrect factoring or sign errors.
  • CThis is the coefficient of x with sign reversed, not a solution to the equation.
Question 16Easy

What is the solution to (x−4)(x+7)=0(x - 4) (x + 7) = 0?

Show the answer and explanation

Why C is right

By the zero product property, either x−4=0x - 4 = 0 or x+7=0x + 7 = 0. Solving the first equation yields x=4x = 4, and solving the second yields x=−7x = -7. Therefore, both -7 and 4 are solutions.

Why the others are wrong

  • AThis is only one of the two solutions; the equation also has the solution x = 4.
  • BThis is only one of the two solutions; the equation also has the solution x = -7.
  • DThese values have the incorrect signs; setting x - 4 = 0 yields x = 4, not x = -4.
Question 17Easy

What is the negative solution to x2−25=0x^{2} - 25 = 0?

Show the answer and explanation

Why B is right

The equation x2x^{2} - 25 = 0 can be rewritten as x2x^{2} = 25. Taking the square root of both sides yields x=5x = 5 or x=−5x = -5. Since the question asks for the negative solution, the answer is -5.

Why the others are wrong

  • AThis is the positive solution to the equation, not the negative one.
  • CThis does not satisfy the original equation.
  • DThis results from confusing x² = 25 with x = 25.
Question 18Easy

What is the positive solution to x2−64=0x^{2} - 64 = 0?

Show the answer and explanation

Why A is right

The equation x2x^{2} - 64 = 0 can be rewritten as x2x^{2} = 64. Taking the square root of both sides yields x=8x = 8 or x=−8x = -8. The positive solution is 8.

Why the others are wrong

  • BThis results from confusing the square root operation, possibly finding the square root of the square root.
  • CThis results from halving 64 instead of taking its square root.
  • DThis is the negative solution, not the positive solution requested.
Question 19Easy

What is the product of the solutions to (x+6)(x−2)=0(x + 6) (x - 2) = 0?

Show the answer and explanation

Why A is right

Setting each factor equal to zero gives x+6=0x + 6 = 0 or x−2=0x - 2 = 0. Solving these equations yields x=−6x = -6 or x=2x = 2. The product of the solutions is (-6)(2) = -12.

Why the others are wrong

  • BThis is the sum of the solutions (-6 + 2 = -4) incorrectly computed, not the product.
  • CThis has the correct magnitude but the wrong sign for the product.
  • DThis accounts for only one of the two solutions and ignores x = 2.
Question 20Easy

What is the product of the solutions to x2+7x+12=0x^{2} + 7x + 12 = 0?

Show the answer and explanation

Why D is right

The equation x2x^{2} + 7x+12=07x + 12 = 0 factors as (x+4)(x+3)=0(x + 4) (x + 3) = 0, giving solutions x=−4x = -4 and x=−3x = -3. The product of the solutions is (-4)(-3) = 12.

Why the others are wrong

  • AThis results from incorrectly computing the product with the wrong sign.
  • BThis represents only one of the two solutions rather than their product.
  • CThis is the coefficient of x, not the product of the solutions.

What to do after easy

Getting these right quickly is the point: speed on the easy questions is what buys time for the hard ones. When they stop costing you effort, move up.

Medium Nonlinear equations · one variable questionsHard Nonlinear equations · one variable questionsStart free →

More on Nonlinear equations · one variable

Questions are written by SAT Climb and drawn from its own item bank. SAT® is a registered trademark of College Board, which is not affiliated with and does not endorse SAT Climb.