Problem 74 - Algebra 2

Question

For what values of \(x\) and \(y\) does \(\sqrt{x^{2} y^{4}}\) equal \(\sqrt[3]{x^{3}} y^{6} ?\) Explain your answer.

Step-by-Step Solution

Answer

The solutions for the equation are \(x=0\) for any real number \(y\), \(x=y^{4}\) when \(y=\sqrt[4]{x}\), and \(x=-y^{4}\) when \(y=-\sqrt[4]{x}\).

Step 1: Understanding the equation

First we need to note down the given equation \(\sqrt{x^{2} y^{4}} = \sqrt[3]{x^{3}} y^{6} \). Here, we have to find the conditions for \(x\) and \(y\) when both sides of the equation are equal.

Step 2: Simplifying Radicals

On left side, \(\sqrt{x^{2} y^{4}}\) can be simplified into \(|x|y^{2}\) as the square root of \(x^{2}\) is the absolute value of \(x\) and square root of \(y^{4}\) is \(y^{2}\). On the right side, \(\sqrt[3]{x^{3}}\) simplifies to \(x\). So the equation simplifies into \( |x|y^{2}=xy^{6} \).

Step 3: Get Rid Of The Mod

We have to consider the two cases for \(|x|\). Either \(x\) can be positive, then \(x\) equals \(|x|\) and we have equation \(x=(xy^{4})\), or \(x\) can be negative, then \(-x\) equals \(|x|\) and we have equation \(-x=(xy^{4})\). From \(x=(xy^{4})\) we get that \(x\) has to be either 0 or \(y^{4}\). From \(-x=(xy^{4})\) we have to exclude \(x=0\) and calculate the values for \(y^{4}\).

Step 4: Conclusion

In conclusion, for \(x=0\), \(y\) can be any real number. For \(x=y^{4}\), \(y\) also has to be \(\sqrt[4]{x}\). For \(x=-y^{4}\), \(y\) has to be \(-\sqrt[4]{x}\).

Key Concepts

Simplifying Radicals

Radicals involve expressions that include roots, such as square roots or cube roots. Simplifying radicals **aims to** make these expressions more manageable.
To simplify, one common approach is to make the numbers under the root sign as small as possible without changing the overall value.

• For instance, \( \sqrt{x^2 y^4} \) can be simplified based on the property \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) if both \( a \) and \( b \) are non-negative.
• We rewrite the expression using exponent rules. \( \sqrt{x^2 y^4} \) breaks down to \( |x|y^2 \), since \( \sqrt{x^2} = |x| \) and \( \sqrt{y^4} = y^2 \).

This simplification process helps equate the expression with other sides of an equation, crucial for solving problems.

Absolute Value

Absolute value represents the non-negative magnitude of a real number, denoted as \( |x| \).
At its core, absolute value strips away any negative sign, implying the distance from zero on a number line.

• For the equation \( \sqrt{x^2 y^4} = \sqrt[3]{x^3} y^6 \), converting \( \sqrt{x^2} \) into \( |x| \) helps in managing sign issues.
• In cases where \( x \) can be positive or negative, absolute value plays a center role in expressing \( x \) without ambiguity.

Overall, using absolute value is necessary when simplifying or solving equations that involve squares or other even powers.

Equating Expressions

Equating expressions refers to the method of determining when two mathematical expressions are the same. This often involves simplifying each side and finding common terms.
For our problem, equating \( |x|y^2 = xy^6 \) derives from simplifying both radical terms involved.

• The absolute value ensures both sides reflect any sign changes in \( x \).
• After simplification, comparing expressions allows us to solve for unknowns: \( x = 0 \) or \( x = y^4 \); and \( x = -y^4 \) introduces additional conditions.

Understanding equating expressions aids in identifying solutions with necessary conditions or constraints, especially when variables possess varied values or forms.

Exponents

Exponents represent repeated multiplication of a base. They play a significant role in simplifying and solving equations with radicals.
To handle radicals effectively, it's necessary to understand and manipulate exponents.

• For example, the cube root \( \sqrt[3]{x^3} \) simplifies naturally to \( x \), illustrating basic exponent rules.
• Additionally, expressions like \( y^6 \) show how powers combine and simplify, providing clarity on the progression of values.

By studying how exponents work within radicals, solving complex equations becomes more intuitive.