Solve x - Square Root of x = Square Root of x: Finding All Values

Q: Why do I need to isolate the square root first?

Isolating $ \sqrt{x} $ gives you one clean radical on each side. This makes squaring both sides much simpler and avoids messy algebra with multiple square root terms.

Q: What happens if I square both sides without isolating first?

You'd get $ (x - \sqrt{x})^2 = (\sqrt{x})^2 $, which expands to $ x^2 - 2x\sqrt{x} + x = x $. Now you still have a square root term to deal with!

Q: Why can't x be negative in this problem?

The expression $ \sqrt{x} $ requires x ≥ 0 for real numbers. Negative values under a square root aren't defined in basic algebra, so we only consider non-negative solutions.

Q: How do I know both solutions are actually correct?

Always substitute back into the original equation:For x = 0: $ 0 - \sqrt{0} = \sqrt{0} $ becomes 0 = 0 ✓For x = 4: $ 4 - \sqrt{4} = \sqrt{4} $ becomes 2 = 2 ✓

Q: Could there be other methods to solve this equation?

You could also substitute $ u = \sqrt{x} $ to get $ u^2 - u = u $, then solve $ u^2 = 2u $. This gives the same result but with different variable names!

Radical Equations with Square Root Isolation

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

What are the possible values of X?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the X values that satisfy the equation

00:03 Collect terms

00:20 Break down X into X squared root

00:26 Find the common factor, and take out of parentheses

00:38 The right side of the equation is 0 because of the subtraction we did

00:42 Find what makes each factor equal zero in the multiplication

00:52 Solve the first option first, square it

00:57 This is one solution

01:01 Now let's solve the second option, isolate X

01:05 Square it to find the solution

01:10 And this is the second possible solution

01:13 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

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

What are the possible values of X?

Step-by-step solution

Let's solve the equation $x - \sqrt{x} = \sqrt{x}$ .

Step 1: Start by simplifying the equation.
The equation is given by:

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

Step 2: Isolate the square root by adding $\sqrt{x}$ to both sides:

$x = 2\sqrt{x}$

Step 3: Square both sides to eliminate the square root.

$(x)^2 = (2\sqrt{x})^2$

This simplifies to:

$x^2 = 4x$

Step 4: Simplify and solve the quadratic equation:

Move all terms to one side:

$x^2 - 4x = 0$

Step 5: Factor the quadratic equation:

$x(x - 4) = 0$

Step 6: Solve for $x$ to find potential solutions:

$x = 0$ or $x = 4$ .

Step 7: Check solutions by substituting back into the original equation:

For $x = 0$ :

$0 - \sqrt{0} = \sqrt{0}$ which simplifies to $0 = 0$ . True.

For $x = 4$ :

$4 - \sqrt{4} = \sqrt{4}$ which simplifies to $4 - 2 = 2$ . True.

Therefore, both solutions are valid.

The possible values of $x$ are 0 and 4.

Therefore, the solution to the problem is $0, 4$ . Thus, the correct choice is option 2.

Final Answer

0, 4

Key Points to Remember

Essential concepts to master this topic

Domain: Square roots require non-negative values under radical
Technique: Isolate $\sqrt{x}$ to get $x = 2\sqrt{x}$ , then square both sides
Check: Substitute x = 0 and x = 4: both give 0 = 0 and 2 = 2 ✓

Common Mistakes

Avoid these frequent errors

Forgetting to check solutions after squaring
Don't assume all solutions from $x^2 = 4x$ are valid = potential extraneous solutions! Squaring both sides can introduce false solutions that don't work in the original equation. Always substitute each solution back into $x - \sqrt{x} = \sqrt{x}$ to verify.

Practice Quiz

Test your knowledge with interactive questions

$$ (4b-3)(4b-3) $$

Rewrite the above expression as an exponential summation expression:

FAQ

Everything you need to know about this question

Why do I need to isolate the square root first?

Isolating $\sqrt{x}$ gives you one clean radical on each side. This makes squaring both sides much simpler and avoids messy algebra with multiple square root terms.

What happens if I square both sides without isolating first?

You'd get $(x - \sqrt{x})^2 = (\sqrt{x})^2$ , which expands to $x^2 - 2x\sqrt{x} + x = x$ . Now you still have a square root term to deal with!

Why can't x be negative in this problem?

The expression $\sqrt{x}$ requires x ≥ 0 for real numbers. Negative values under a square root aren't defined in basic algebra, so we only consider non-negative solutions.

How do I know both solutions are actually correct?

Always substitute back into the original equation:

For x = 0: $0 - \sqrt{0} = \sqrt{0}$ becomes 0 = 0 ✓
For x = 4: $4 - \sqrt{4} = \sqrt{4}$ becomes 2 = 2 ✓

Could there be other methods to solve this equation?

You could also substitute $u = \sqrt{x}$ to get $u^2 - u = u$ , then solve $u^2 = 2u$ . This gives the same result but with different variable names!

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