Simplifying Radicals and Combining Like Radicals - College Algebra

1. Which expression is equivalent to $\sqrt{72x^3y^5}$ when simplified completely, assuming all variables represent positive real numbers?
   Answer: $6xy^2\sqrt{2xy}$ To simplify $\sqrt{72x^3y^5}$, first factor out perfect squares: $72 = 36 \cdot 2 = 6^2 \cdot 2$, $x^3 = x^2 \cdot x$, and $y^5 = y^4 \cdot y = (y^2)^2 \cdot y$. So $\sqrt{72x^3y^5} = \sqrt{6^2 \cdot 2 \cdot x^2 \cdot x \cdot (y^2)^2 \cdot y} = 6 \cdot x \cdot y^2 \sqrt{2xy} = 6xy^2\sqrt{2xy}$. Choice B incorrectly factors the exponents as $x^3 = x^2 \cdot x$ but writes $x^2$ outside. Choice C misses the $y$ under the radical. Choice D incorrectly takes $\sqrt{36} = 36$.
2. Simplify $\sqrt{8x^2} + x\sqrt{18} - 2\sqrt{2x^2}$ where $x > 0$.
   Answer: $3x\sqrt{2}$ Since $x > 0$, we have $\sqrt{x^2} = x$. Simplifying each term: $\sqrt{8x^2} = \sqrt{4 \cdot 2 \cdot x^2} = 2x\sqrt{2}$, $x\sqrt{18} = x\sqrt{9 \cdot 2} = 3x\sqrt{2}$, and $2\sqrt{2x^2} = 2x\sqrt{2}$. Combining like terms: $2x\sqrt{2} + 3x\sqrt{2} - 2x\sqrt{2} = (2 + 3 - 2)x\sqrt{2} = 3x\sqrt{2}$. Choice A results from incorrectly combining coefficients. Choice D shows the expression before combining like terms. Choice C adds all coefficients without considering the subtraction.
3. If $\sqrt{a} + \sqrt{b} = 7$ and $\sqrt{a} - \sqrt{b} = 1$, what is the value of $ab$?
   Answer: $144$ Let $x = \sqrt{a}$ and $y = \sqrt{b}$. Then we have the system: $x + y = 7$ and $x - y = 1$. Adding these equations: $2x = 8$, so $x = 4$, which means $\sqrt{a} = 4$ and $a = 16$. Substituting back: $4 + y = 7$, so $y = 3$, which means $\sqrt{b} = 3$ and $b = 9$. Therefore, $ab = 16 \cdot 9 = 144$. Choice A represents $a + b$. Choice B represents just $a$. Choice D represents $(\sqrt{a} + \sqrt{b})^2$ without expanding correctly.
4. Rationalize the denominator and simplify: $\frac{4}{\sqrt{12} + \sqrt{3}}$
   Answer: $\frac{4\sqrt{3}}{9}$ First simplify $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$. So the expression becomes $\frac{4}{2\sqrt{3} + \sqrt{3}} = \frac{4}{3\sqrt{3}}$. To rationalize, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$: $\frac{4}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3 \cdot 3} = \frac{4\sqrt{3}}{9}$. Choice B incorrectly uses the conjugate method on the simplified form. Choice C has the wrong denominator (should be 9, not 3). Choice D is the intermediate step before rationalization.

Start this quiz