SAT Math · Advanced Math

Nonlinear equations and systems on the SAT

Nonlinear equations include square roots, absolute values, variables in a denominator, and systems where a line meets a parabola. Isolate the tricky part, then undo it: square both sides, split into two cases, or multiply by the denominator. Always check every answer in the original equation, because squaring and clearing denominators can create extraneous solutions that do not really work.

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The key idea

How to undo each kind
Equation hasMoveWatch out for
A square rootIsolate the root, then square both sidesAnswers that make the root equal a negative number
An absolute valueIsolate it, then split into ++ and −- cases∣A∣=|A| = a negative number has no solution
x in a denominatorMultiply every term by the denominatorAnswers that make a denominator 0
A line and a parabolaSet the y's equal, solve the quadratic0, 1 or 2 intersection points

Squaring is not reversible: 323^2 and (−3)2(-3)^2 are both 9. So after squaring you may pick up an answer that fit the squared equation but not the original. That is an extraneous solution, and the SAT loves to offer it as a choice.

For a line-parabola system, substitution leads to a quadratic. Its discriminant counts the intersections:

b2−4ac>0: two pointsb2−4ac=0: one pointb2−4ac<0: noneb^2 - 4ac > 0: \text{ two points} \qquad b^2 - 4ac = 0: \text{ one point} \qquad b^2 - 4ac < 0: \text{ none}

Worked examples

Example 1: a radical equation with an extraneous solution

Problem Solve x+7=x−5\sqrt{x + 7} = x - 5.

  1. The root is already alone. Square both sides.
    x+7=x2−10x+25x + 7 = x^2 - 10x + 25
  2. Move everything to one side.
    x2−11x+18=0x^2 - 11x + 18 = 0
  3. Factor.
    (x−9)(x−2)=0(x - 9)(x - 2) = 0
  4. Check x=9x = 9: 16=4\sqrt{16} = 4 and 9−5=49 - 5 = 4. It works.
    9+7=9−5\sqrt{9 + 7} = 9 - 5
  5. Check x=2x = 2: 9=3\sqrt{9} = 3, but 2−5=−32 - 5 = -3. It fails, so 2 is extraneous.

Answer x=9x = 9

Example 2: absolute value

Problem Solve ∣2x−3∣=7|2x - 3| = 7.

  1. The inside is either 7 or −7-7. First case:
    2x−3=72x - 3 = 7
  2. Second case:
    2x−3=−72x - 3 = -7
  3. Solve each: 2x=102x = 10 gives x=5x = 5, and 2x=−42x = -4 gives x=−2x = -2. Both check: ∣7∣=7|7| = 7 and ∣−7∣=7|-7| = 7.

Answer x=5x = 5 or x=−2x = -2

Example 3: a rational equation with no solution

Problem Solve xx−4+2=4x−4\frac{x}{x - 4} + 2 = \frac{4}{x - 4}.

  1. Multiply every term by x−4x - 4, including the 2.
    x+2(x−4)=4x + 2(x - 4) = 4
  2. Simplify.
    3x−8=43x - 8 = 4
  3. Solve.
    x=4x = 4
  4. But x=4x = 4 makes the denominator x−4x - 4 equal 0, so it is not allowed. It is extraneous, and there is nothing else.

Answer No solution

Example 4 (SAT-hard): a line tangent to a parabola

Problem The system y=x2−4x+1y = x^2 - 4x + 1 and y=2x+ky = 2x + k has exactly one solution. What is the value of k?

  1. Set the two expressions for y equal and move everything to one side.
    x2−6x+(1−k)=0x^2 - 6x + (1 - k) = 0
  2. Exactly one solution means the discriminant is 0.
    36−4(1−k)=036 - 4(1 - k) = 0
  3. Divide by 4: 9−(1−k)=09 - (1 - k) = 0, so 8+k=08 + k = 0.
    k=−8k = -8
  4. Check: x2−6x+9=(x−3)2x^2 - 6x + 9 = (x - 3)^2, so the line touches the parabola only at x=3x = 3, the point (3,−2)(3, -2).

Answer k=−8k = -8

Common mistakes

  • Squaring before isolating the root. Squaring x+2=5\sqrt{x} + 2 = 5 as is creates a messy middle term. Fix: subtract 2 first, then square.
  • Squaring x−5x - 5 as x2−25x^2 - 25. (x−5)2=x2−10x+25(x - 5)^2 = x^2 - 10x + 25. Fix: write it as (x−5)(x−5)(x - 5)(x - 5).
  • Keeping both answers without checking. Squaring can add a fake answer. Fix: plug each answer into the original equation.
  • Solving only the positive case. ∣2x−3∣=7|2x - 3| = 7 has two cases. Fix: always write both =7= 7 and =−7= -7.
  • Forgetting that a denominator cannot be 0. If your answer makes any denominator 0, throw it out.

Quick method

Practice

5 SAT-style questions

  1. What is the solution to 2x+1=5\sqrt{2x + 1} = 5?

    1. 22
    2. 1212
    3. 1313
    4. 2424
    Show answer

    Answer: 1212

    Square both sides: 2x+1=252x + 1 = 25, so 2x=242x = 24 and x=12x = 12. Check: 25=5\sqrt{25} = 5. 2 comes from 2x+1=52x + 1 = 5, which forgets to square the 5. 13 adds the 1 instead of subtracting it, and 24 forgets to divide by 2.

  2. What are all solutions to ∣x+4∣=9|x + 4| = 9?

    1. 55 and −13-13
    2. 55 only
    3. −5-5 and 1313
    4. 55 and 1313
    Show answer

    Answer: 55 and −13-13

    Either x+4=9x + 4 = 9, giving 5, or x+4=−9x + 4 = -9, giving −13-13. "5 only" forgets the negative case. −5-5 and 13 solve ∣x−4∣=9|x - 4| = 9 instead.

  3. What are all solutions to x+2=x\sqrt{x + 2} = x?

    1. −1-1 and 22
    2. 22 only
    3. −1-1 only
    4. No solution
    Show answer

    Answer: 22 only

    Squaring gives x2−x−2=0x^2 - x - 2 = 0, so x=2x = 2 or x=−1x = -1. Check −1-1: 1=1\sqrt{1} = 1, not −1-1. It is extraneous, so only 2 works.

  4. Which ordered pair is a solution to the system y=x2−1y = x^2 - 1 and y=x+1y = x + 1?

    1. (1,0)(1, 0)
    2. (2,3)(2, 3)
    3. (−2,3)(-2, 3)
    4. (0,1)(0, 1)
    Show answer

    Answer: (2,3)(2, 3)

    Set x2−1=x+1x^2 - 1 = x + 1: x2−x−2=0x^2 - x - 2 = 0, so x=2x = 2 or x=−1x = -1. The points are (2,3)(2, 3) and (−1,0)(-1, 0). The other choices each fit at most one of the equations.

  5. Student-produced response: what is the positive solution to 12x=x+1\frac{12}{x} = x + 1?

    Show answer

    Answer: 3

    Multiply by x: 12=x2+x12 = x^2 + x, so x2+x−12=0x^2 + x - 12 = 0 and (x+4)(x−3)=0(x + 4)(x - 3) = 0. The solutions are −4-4 and 3, and the positive one is 3. Check: 123=4=3+1\frac{12}{3} = 4 = 3 + 1.

Frequently asked questions

What is an extraneous solution?

It is an answer that your algebra produced but that does not work in the original equation. It shows up after squaring both sides or multiplying by an expression with x. That is why every answer to a radical or rational equation must be checked.

Can an absolute value equation have no solution?

Yes. An absolute value is never negative, so ∣x−2∣=−5|x - 2| = -5 has no solution. If the equation is ∣x−2∣+8=3|x - 2| + 8 = 3, isolate the absolute value first: ∣x−2∣=−5|x - 2| = -5, and stop there.

How many times can a line and a parabola meet?

Zero, one or two times. Substitute to get a quadratic and look at its discriminant: positive means two points, zero means the line just touches (tangent), and negative means they never meet.

Sources

  1. College Board: SAT Math, Advanced Math skills (nonlinear equations, systems in 2 variables), accessed October 1, 2026

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