Question

algebra 2 quiz review name:
1 simplify the expression
\\(\sqrt6{q^{12}z^4}\\)
2 find the reduced radical form of
\\(\sqrt3{250y^3z^4}\\)
3 find the reduced radical form of
\\(\frac{7}{1 - \sqrt{6}}\\)
4 find the reduced radical form of
\\(\frac{3}{1 - \sqrt{5}}\\)
5 \\(f(x) = \sqrt{x - 2} - 3\\)
8 \\(f(x) = \sqrt{x + 3} - 1\\)

Explanation:

Problem 1: Simplify $\boldsymbol{\sqrt[4]{q^{12}z^4}}$

Step 1: Apply the nth root property

For a radical $\sqrt[n]{a^m}$, we can simplify it as $a^{\frac{m}{n}}$ (when $a\geq0$ for even roots). So we apply this to $q^{12}$ and $z^4$ separately.
For $q^{12}$: $\sqrt[4]{q^{12}} = q^{\frac{12}{4}} = q^3$
For $z^4$: $\sqrt[4]{z^4} = z^{\frac{4}{4}} = z$ (assuming $z\geq0$ to take the real root)

Step 2: Combine the results

Multiply the simplified parts together: $q^3 \cdot z = q^3z$

Step 1: Factor the radicand

Factor 250: $250 = 125 \times 2 = 5^3 \times 2$
Factor $y^3$: $y^3 = y^3$
Factor $z^4$: $z^4 = z^3 \times z$
So the radicand becomes $5^3 \times 2 \times y^3 \times z^3 \times z$

Step 2: Apply the cube root property

$\sqrt[3]{a^3b} = a\sqrt[3]{b}$. So we take the cube root of the perfect cube factors:
$\sqrt[3]{5^3y^3z^3} = 5yz$
The remaining factor is $2z$, so we have:
$\sqrt[3]{250y^3z^4} = 5yz\sqrt[3]{2z}$

Step 1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator, which is $1 + \sqrt{6}$.
$\frac{7(1 + \sqrt{6})}{(1 - \sqrt{6})(1 + \sqrt{6})}$

Step 2: Simplify the denominator

The denominator is a difference of squares: $(a - b)(a + b) = a^2 - b^2$. So:
$(1)^2 - (\sqrt{6})^2 = 1 - 6 = -5$

Step 3: Simplify the numerator

Multiply out the numerator: $7(1 + \sqrt{6}) = 7 + 7\sqrt{6}$

Step 4: Combine and simplify

Now we have $\frac{7 + 7\sqrt{6}}{-5} = -\frac{7 + 7\sqrt{6}}{5}$ or we can factor out 7: $-\frac{7(1 + \sqrt{6})}{5}$

Answer:

$q^3z$

Problem 2: Simplify $\boldsymbol{\sqrt[3]{250y^3z^4}}$