What It Targets

• Radical Equations
• Square Roots
• Advanced Math
• Solution Verification
• SAT Trap Questions

The Core Idea

Whenever you solve a radical equation, there is one final step:

CHECK YOUR ANSWER.

The SAT loves radical equations because they can create:

Extraneous Solutions

These are fake answers that appear during your algebra but do not actually work in the original equation.

What Is an Extraneous Solution?

An extraneous solution is:

A solution that appears after your algebra steps but does not satisfy the original equation.

Think of it as:

A fake answer created by the solving process.

Why Do Extraneous Solutions Happen?

They usually happen when we:

Square both sides.

Squaring is not a perfectly reversible operation.

For example:

If we start with:

$1 = -1$

we know this isn’t true. But if I decided to square both sides, just because I can (and we are following our reciprocity rule of “what we do to one side, we must do to the other”)

$1^2=(-1)^2$

Both become:

$1$ (i.e. 1 = 1)

which makes it seem like the equation is true for whatever “solution” we tried, when it isn’t. The squaring process erased important information. As a result, squaring can accidentally create extra answers that were never valid in the original problem.

The SAT Rule

Whenever you see:

• $\sqrt{x}$
• $\sqrt{x+3}$
• $\sqrt{2x-5}$

and you square both sides,

you MUST check every answer in the original equation.

Not the simplified equation.

The original equation.

Quick Example

Solve:

$\sqrt{x+2}=x$

Square both sides:

$x+2=x^2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

Potential solutions:

$x=2$

$x=-1$

Now check them in the original equation.

Check $x=2$:

$\sqrt{2+2}=2$

$2=2$ ✅

Check $x=-1$:

$\sqrt{-1+2}=-1$

$1=-1$ ❌ false.

Therefore:

Valid Solution:

$x=2$

Extraneous Solution:

$x=-1$

SAT Shortcut

Before checking anything, ask:

Can the right side be negative?

Example:

$\sqrt{x-3}=5-x$

The square root side is always positive or zero.

That means:

$5-x$

must also be positive or zero.

Immediately:

$x\leq5$

Any solution larger than 5 is suspicious.

Checkpoint 1

Is $x=7$ likely to work in:

$\sqrt{x-3}=5-x$

Answer

No.

Because:

$5-7=-2$

A square root cannot equal a negative number.

Another SAT Shortcut

A principal square root is NEVER negative.

Examples:

$\sqrt{25}=5$

NOT

$\sqrt{25}=-5$

Students lose points on this constantly because they are confusing it with a different question:

$x^2=25$

In that equation, we are asking:

"What numbers can be squared to produce 25?"

Answer:

$x=5$ OR $x=-5$

because: $5^2=25$ AND $(-5)^2=25$

However, when you see:

$\sqrt{25}$

the square root symbol is asking for the principal (positive) square root only.

Therefore:

$\sqrt{25}=5$

not

$\pm5$

and not

• $-5$

Checkpoint 2

True or False:

$\sqrt{16}=-4$

Answer

False.

$\sqrt{16}=4$

Guided Practice

Question 1

What is the complete solution set for:

$\sqrt{x+2}=x$

Answer

$x=2$

Why?

Square both sides:

$x+2=x^2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

Potential solutions:

$x=2$

$x=-1$

Check:

$x=-1$

$\sqrt{1}=1\neq-1$

Extraneous.

Question 2

Solve:

$\sqrt{2x+7}=x+2$

Identify any extraneous solutions.

Answer

Valid:

$x=1$

Extraneous:

$x=-3$

Why?

Square both sides:

$2x+7=x^2+4x+4$

$x^2+2x-3=0$

$(x+3)(x-1)=0$

Check both answers.

Only $x=1$ works.

Question 3

If

$\sqrt{x-3}=5-x$

how many valid real solutions exist?

Answer

Exactly one.

$x=4$

Why?

Square:

$x-3=25-10x+x^2$

$x^2-11x+28=0$

$(x-7)(x-4)=0$

Potential solutions:

$x=7$

$x=4$

Check:

$x=7$

$\sqrt4=-2$

False.

$x=4$

$\sqrt1=1$

True.

Drill

1

Solve for $x$:

$\sqrt{3x+13}=x+3$

2

Given

$\sqrt{4x+1}=x-1$

why is $x=0$ considered an extraneous solution?

3

Solve:

$\sqrt{x+14}=x-2$

4

For the equation

$\sqrt{2x^2-2}=x+1$

what are all valid real values of $x$?

5

Solve:

$\sqrt{5x-1}=x+1$

6

True or False:

Squaring both sides of an equation can eliminate valid solutions.

7

What value of $x$ satisfies:

$\sqrt{x-1}+3=1$

Answer Key

1

$x=1$

Square:

$3x+13=x^2+6x+9$

$x^2+3x-4=0$

$(x+4)(x-1)=0$

Check:

$x=-4$

$\sqrt1=-1$

False.

$x=1$

Works.

2

Because it creates:

$\sqrt1=-1$

A principal square root can never equal a negative number.

Therefore $x=0$ is extraneous.

3

$x=5$

Square:

$x+14=x^2-4x+4$

$x^2-5x-10=0$

Potential solutions:

$x=5$

$x=-2$

Check:

$x=-2$

$\sqrt{12}=-4$

False.

4

$x=3$

and

$x=-1$

Square:

$2x^2-2=x^2+2x+1$

$x^2-2x-3=0$

$(x-3)(x+1)=0$

Both solutions satisfy the original equation.

5

$x=1$

or

$x=2$

Square:

$5x-1=x^2+2x+1$

$x^2-3x+2=0$

$(x-1)(x-2)=0$

Both check correctly.

6

False.

Squaring can introduce extraneous solutions.

It does not eliminate valid solutions.

7

No real solution.

Rearrange:

$\sqrt{x-1}=-2$

A square root can never equal a negative number.

Therefore there is no real solution.

SAT Shortcut Summary

Whenever you solve a radical equation:

1. Isolate the radical.
2. Square both sides.
3. Solve.
4. CHECK EVERY ANSWER in the original equation.

Remember:

Extraneous solutions are fake answers created by squaring.

The SAT loves them.

If you forget to check, you will eventually lose points.