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Advanced Math / LESSON 17

Keep valid solutions after nonlinear steps · PSAT/NMSQT Math

10 min + practice · Original foundation lesson

Squaring and absolute values change how equations encode signs. Candidate roots must satisfy the original equation and any contextual restrictions.

Choose a representation

Before squaring a square-root equation, require its other side to be nonnegative. Use two signed branches for a positive absolute-value distance. Use quadratic structure when a sum or height is requested.

Worked example

What is the real solution of √(x+12)=x? The right side requires x≥0 before squaring. Squaring gives (x−4)(x+3)=0; reject the negative candidate in the original equation. Check x=4: √(4+12)=4; candidate -3 fails because a square root cannot equal that negative value. The requested value is 4.

Contrast a wrong operation

In that example, -3 follows this error: Keep the extraneous negative candidate introduced by squaring. The correct requested value is 4. Explain which quantity the mistaken method is actually calculating.

Connect a second form

For 3 x²−12 x+6=0, let r and s be its roots, counted with multiplicity. What is r+s? Use root sum equal to negative linear coefficient divided by leading coefficient. The nonmonic leading coefficient must remain in the denominator. Substitute the given numbers: 12 ÷ 3 = 4. The requested value is 4.

Transfer and check

What is the real solution of √(x+20)=x? Pause before checking. Answer: 5. The right side requires x≥0 before squaring. Squaring gives (x−5)(x+4)=0; reject the negative candidate in the original equation. Check x=5: √(5+20)=5; candidate -4 fails because a square root cannot equal that negative value.

Boundaries of the method

Squaring can introduce an extraneous root. The root sum of a nonmonic quadratic divides by its leading coefficient. A maximum height is an output, not the time when it occurs. This lesson samples a PSAT/NMSQT objective; its examples are original and do not claim official question status or full coverage.

WORKED EXAMPLE

What is the real solution of √(x+12)=x?

  1. The right side requires x≥0 before squaring.
  2. Squaring gives (x−4)(x+3)=0; reject the negative candidate in the original equation.
  3. Check x=4: √(4+12)=4; candidate -3 fails because a square root cannot equal that negative value.
  4. The requested value is 4.
Answer: 4
Compare the other answer choices
12

Report the constant under the radical as the solution.

16

Report the square of the valid input rather than the input.

-3

Keep the extraneous negative candidate introduced by squaring.

Try it, then explain it.

Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.

Practice this skill

Sources & authorship

Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.

Math specifications
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  • The reasoning target
  • Choose a representation
  • Worked example