Solve for X: Square Root Equation sqrt(2)·sqrt(4)/sqrt(16) = x/sqrt(8)

Square Root Simplification with Cross-Multiplication

Solve for x:

2416=x8 \frac{\sqrt{2}\cdot\sqrt{4}}{\sqrt{16}}=\frac{x}{\sqrt{8}}

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:09 Let's find the value of X together.
00:12 First, we'll isolate X to solve the equation.
00:29 Now, use the substitution rule and rearrange.
00:34 Remember, multiplying the root of A by the root of B
00:38 gives you the root of A times B. Easy, right?
00:44 Let's apply this rule and find the product.
00:48 Try to simplify wherever you can.
00:52 For example, break down 4 into 2 squared.
00:57 And there you have it! That's the solution.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve for x:

2416=x8 \frac{\sqrt{2}\cdot\sqrt{4}}{\sqrt{16}}=\frac{x}{\sqrt{8}}

2

Step-by-step solution

Let's solve the equation step by step:

  • Step 1: Simplify the left side of the equation
    We are given 2416 \frac{\sqrt{2}\cdot\sqrt{4}}{\sqrt{16}} . Start by simplifying each square root:
    • 2 \sqrt{2} remains as it is.
    • 4=2 \sqrt{4} = 2 because 4=22 4 = 2^2 .
    • 16=4 \sqrt{16} = 4 because 16=42 16 = 4^2 .
    Substitute these into the initial expression: 224=224 \frac{\sqrt{2} \cdot 2}{4} = \frac{2\sqrt{2}}{4} .
  • Step 2: Simplify the expression
    The expression 224 \frac{2\sqrt{2}}{4} simplifies to 22 \frac{\sqrt{2}}{2} .
  • Step 3: Equate to the right side
    Set this equal to the right side: 22=x8 \frac{\sqrt{2}}{2} = \frac{x}{\sqrt{8}} .
  • Step 4: Cross-multiply to solve for x x
    Cross-multiplying gives x2=28 x \cdot 2 = \sqrt{2} \cdot \sqrt{8} . Simplify the right side:
    • 8=42=42=22 \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}
    • So, 28=222=2(22)=22=4 \sqrt{2} \cdot \sqrt{8} = \sqrt{2} \cdot 2\sqrt{2} = 2 \cdot (\sqrt{2} \cdot \sqrt{2}) = 2 \cdot 2 = 4
    • The equation becomes 2x=4 2x = 4 .
    • Solving for x x , divide both sides by 2: x=42=2 x = \frac{4}{2} = 2 .

Therefore, the solution to the problem is x=2 x = 2 .

3

Final Answer

2 2

Key Points to Remember

Essential concepts to master this topic
  • Simplification: Evaluate each square root to its simplest form first
  • Technique: 4=2 \sqrt{4} = 2 and 16=4 \sqrt{16} = 4 before multiplying or dividing
  • Verification: Substitute x = 2 back: 22=28 \frac{\sqrt{2}}{2} = \frac{2}{\sqrt{8}} both equal 22 \frac{\sqrt{2}}{2}

Common Mistakes

Avoid these frequent errors
  • Working with unsimplified square roots throughout
    Don't leave 4 \sqrt{4} and 16 \sqrt{16} as radicals = messy fractions like 2416 \frac{\sqrt{2}\cdot\sqrt{4}}{\sqrt{16}} ! This makes cross-multiplication confusing and leads to calculation errors. Always simplify perfect square roots first: 4=2 \sqrt{4} = 2 and 16=4 \sqrt{16} = 4 .

Practice Quiz

Test your knowledge with interactive questions

Solve the following exercise:

\( \sqrt{\frac{2}{4}}= \)

FAQ

Everything you need to know about this question

Do I have to simplify the square roots first?

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Yes, absolutely! Perfect squares like 4 \sqrt{4} and 16 \sqrt{16} should always be simplified to whole numbers before doing any other operations. This makes the rest much easier!

How do I know when to cross-multiply?

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Cross-multiply when you have one fraction equals another fraction, like ab=cd \frac{a}{b} = \frac{c}{d} . This gives you ad=bc ad = bc , which is usually easier to solve.

What if I can't simplify a square root?

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Some square roots like 2 \sqrt{2} can't be simplified further. That's okay! Just leave them as radicals and work with them algebraically.

Why does √2 · √8 = 4?

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Use the property ab=ab \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} : 28=16=4 \sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4 . Or think of it as 222=2(2)2=22=4 \sqrt{2} \cdot 2\sqrt{2} = 2(\sqrt{2})^2 = 2 \cdot 2 = 4 .

Can I just use decimals instead of radicals?

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While you could use decimals, it's better to keep exact radical form. Decimals like 1.414... for 2 \sqrt{2} are approximations and can lead to rounding errors in your final answer.

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