Week 5: Solving Radical Equations

This week in Pre Calculus 11, I learned how to solve radical equations. It was a little bit difficult for me at some points, because I didn’t understand squaring. Today I will teach you how to deal with these kinds of expressions.

 

Steps: 

  1. Solving
  2. Restrictions
  3. Checking

Example

\sqrt{x} = 6

SOLVING

  • remove the radical sign by squaring both sides of the equation.

(\sqrt{x})^2 = 6^2

  • Simplify

x = 36

Restrictions

  Note: a number under a  radical sign can not be negative because a number times itself is positive weather it be -x or x. The only exception is if it has a odd power.

In the example above x needs to be a positive number so its restriction would be x > 0 or x = 0.

Check It Out

  • \sqrt{36} = 6
  • 6 = 6

This is how to solve a basic radical equation. Now I will move on to a more difficult equation to maximize your understanding.

 

Example 2

1 -3\sqrt{2x} = -3 -2\sqrt{2x}

 Note: When solving eq, what you do to one side you have to do to the other.

SOLVING

  1. Place the constants on side of the equation
  • 1 + 3 -3\sqrt{2x} = -3 + 3 -2\sqrt{2x}
  •  4 -3\sqrt{2x} = -2\sqrt{2x}

2. Combine the radical terms

 Note: The radicans are the same so they can be combined together. Place them on one side of the equation.

  • 4 -3\sqrt{2x} = -2\sqrt{2x}
  • 4 -3\sqrt{2x} + 3\sqrt{2x} = -2\sqrt{2x} + 3\sqrt{2x}
  • 4 = 1\sqrt{2x}

3. Square both sides to remove the radical sign.

  • 4^2(\sqrt{2x})^2
  • 4^2 = 2x
  • 16 = 2x

4. Isolate x

  • \frac{16}{2}\frac{2x}{2}
  • \frac{16}{2} = x
  • 8 = x

RESTRICTIONS

x > 0   or   x = 0

Check It Out

  • 1 – 3\sqrt{2(8)} = -3 – 2\sqrt{2(8)}
  • 1 – 3\sqrt{16} = -3 – 2\sqrt{16}
  • 1 – 3\cdot4 = -3 – 2\cdot4
  • 1 – 12 = -3 – 8
  • -11 = -11

This is how you solve radical equations.

Comment below if this helped you or if you have any questions. 🙂

 

 

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