Question

chapter 1.1a

1. which is equivalent to $49^{\frac{3}{2}}$?

a 21
b 98
c 294
d 343

1. which expression is another way to write $\sqrt{125x^4}$?

a $5x^{\frac{1}{4}}$
b $5x^{\frac{4}{3}}$
c $25x^{\frac{1}{2}}$
d $25x^{\frac{4}{3}}$

1. if x and y are real numbers, what is the simplified radical form of $(x^2y^3)^{\frac{1}{2}}$?

a $y\sqrt{x^2}$
b $y\sqrt{x^3}$
c $|y|\sqrt{x^2}$
d $|y|\sqrt{x^3}$

chapter 1.1b

1. what is the simplified expression of $\sqrt{\frac{9x^4}{4x^{10}}}$?

a $3x$
b $9x$
c $3x^3$
d $9x^3$

1. what is the simplified form of $(2\sqrt{5} + 3)(\sqrt{5} - 1)$?

a $\sqrt{5} - 3$
b $\sqrt{5} + 1$
c $2\sqrt{5} - 3$
d $2\sqrt{5} + 1$

Explanation:

Question 1

Step 1: Recall the exponent rule \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\)

So \(49^{\frac{3}{2}}=\sqrt{49^3}\)

Step 2: Simplify \(\sqrt{49^3}\)

First, \(49 = 7^2\), so \(49^3=(7^2)^3 = 7^{6}\)
Then \(\sqrt{7^{6}}=7^{3}\) (since \(\sqrt{x}=x^{\frac{1}{2}}\), so \((7^{6})^{\frac{1}{2}}=7^{3}\))

Step 3: Calculate \(7^3\)

\(7^3 = 7\times7\times7=343\)

Step 1: Simplify \(\sqrt{125x^4}\)

First, factor 125: \(125 = 5^3=5^2\times5\)
So \(\sqrt{125x^4}=\sqrt{5^2\times5\times x^4}\)

Step 2: Use the square - root property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\))

\(\sqrt{5^2\times5\times x^4}=\sqrt{5^2}\times\sqrt{5}\times\sqrt{x^4}\)
We know that \(\sqrt{5^2} = 5\), \(\sqrt{x^4}=x^{2}\) (since \((x^{2})^2=x^{4}\))
So \(\sqrt{5^2}\times\sqrt{5}\times\sqrt{x^4}=5\times\sqrt{5}\times x^{2}\)? Wait, no, wait. Wait, maybe I made a mistake. Wait, \(125 = 25\times5\), \(25 = 5^2\)
So \(\sqrt{125x^4}=\sqrt{25\times5\times x^4}=\sqrt{25}\times\sqrt{5}\times\sqrt{x^4}\)? No, wait, \(\sqrt{125x^4}=\sqrt{25\times5\times x^4}=\sqrt{25}\times\sqrt{5x^4}\)? No, no, the correct factoring: \(125 = 5^3\), \(x^4=(x^2)^2\)
Wait, \(\sqrt{125x^4}=\sqrt{5^3\times x^4}=\sqrt{5^2\times5\times x^4}\)
\(=\sqrt{5^2}\times\sqrt{5}\times\sqrt{x^4}=5\times x^{2}\times\sqrt{5}\)? No, this is not in the options. Wait, maybe the original problem is \(\sqrt{125x^8}\)? Wait, the option has \(x^{\frac{4}{3}}\) or \(x^{\frac{1}{2}}\). Wait, maybe the problem is \(\sqrt[3]{125x^4}\)? Let's re - examine the options. The options have exponents like \(\frac{1}{2}\), \(\frac{4}{3}\). So maybe it's a cube root. Let's assume it's \(\sqrt[3]{125x^4}\) (maybe a typo in the problem, square root or cube root).
If it's \(\sqrt[3]{125x^4}\):

Step 1: Recall the cube - root rule \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) (\(a,b\in R\))

\(\sqrt[3]{125x^4}=\sqrt[3]{125}\times\sqrt[3]{x^4}\)

Step 2: Simplify \(\sqrt[3]{125}\) and \(\sqrt[3]{x^4}\)

We know that \(125 = 5^3\), so \(\sqrt[3]{125}=5\)
And \(\sqrt[3]{x^4}=\sqrt[3]{x^3\times x}=x\sqrt[3]{x}=x^{1 + \frac{1}{3}}=x^{\frac{4}{3}}\) (using \(a^m\times a^n=a^{m + n}\) and \(\sqrt[3]{x}=x^{\frac{1}{3}}\))
So \(\sqrt[3]{125x^4}=5x^{\frac{4}{3}}\)

Step 1: Apply the power of a power rule \((a^m)^n=a^{mn}\) to \((x^{2}y^{3})^{\frac{1}{2}}\)

\((x^{2}y^{3})^{\frac{1}{2}}=x^{2\times\frac{1}{2}}y^{3\times\frac{1}{2}}=x^{1}y^{\frac{3}{2}}\)

Step 2: Rewrite \(y^{\frac{3}{2}}\) and \(x\) in radical form

\(y^{\frac{3}{2}}=y\sqrt{y}\) (since \(y^{\frac{3}{2}}=y^{1+\frac{1}{2}}=y\times y^{\frac{1}{2}}=y\sqrt{y}\)) and \(x = x\)
But wait, we need to consider the domain. Since \(x\) and \(y\) are real numbers, for \((x^{2}y^{3})^{\frac{1}{2}}\), we have \(x^{2}y^{3}\geq0\). Since \(x^{2}\geq0\) for all real \(x\), then \(y^{3}\geq0\), so \(y\geq0\)
Wait, no, the options have absolute values. Wait, maybe I made a mistake. Wait, \((x^{2}y^{3})^{\frac{1}{2}}=\sqrt{x^{2}y^{3}}\)
\(\sqrt{x^{2}y^{3}}=\sqrt{x^{2}}\times\sqrt{y^{3}}\)
We know that \(\sqrt{x^{2}}=\vert x\vert\) and \(\sqrt{y^{3}}=y\sqrt{y}\) (since \(y^{3}=y^{2}\times y\) and \(y\geq0\) because \(y^{3}\) is under the square root when \(x^{2}y^{3}\geq0\) and \(x^{2}\geq0\))
Wait, the options:
Option A: \(y\sqrt[3]{x^{2}}\) No. Option B: \(y\sqrt{x^{2}}\) No. Option C: \(\vert y\vert\sqrt[3]{x^{2}}\) No. Option D: \(\vert x\vert\sqrt{x^{2}}\)? Wait, no, wait the original problem: \(((x^{2}y^{3})^{\frac{1}{2}})\)? Wait, maybe the original problem is \(((x^{2}y^{6})^{\frac{1}{2}})\)? No, the user's problem is \(((x^{2}y^{3})^{\frac{1}{2}})\)
Wait, let's re - do it. \((x^{2}y^{3})^{\frac{1}{2}}=x^{2\times\frac{1}{2}}y^{3\times\frac{1}{2}}=x y^{\frac{3}{2}}\)
\(y^{\frac{3}{2}}=y\sqrt{y}\), and \(x\) if we consider the square root of \(x^{2}\) is \(\vert x\vert\), but in our case, the exponent of \(x\) is 2, when we take the square root, \(\sqrt{x^{2}}=\vert x\vert\), and the exponent of \(y\) is 3, \(\sqrt{y^{3}}=y\sqrt{y}\) (since \(y^{3}\geq0\) implies \(y\geq0\))
Wait, the options:
A. \(y\sqrt[3]{x^{2}}\)
B. \(y\sqrt{x^{2}}\)
C. \(\vert y\vert\sqrt[3]{x^{2}}\)
D. \(\vert x\vert\sqrt{x^{2}}\)
Wait, maybe there is a typo in the problem. If the problem is \(((x^{2}y^{6})^{\frac{1}{2}})\), then \((x^{2}y^{6})^{\frac{1}{2}}=x y^{3}\), but no. Wait, maybe the original problem is \(((x^{3}y^{2})^{\frac{1}{2}})\)? No. Wait, let's check the options again. Option C is \(\vert y\vert\sqrt[3]{x^{2}}\), Option A is \(y\sqrt[3]{x^{2}}\). Wait, maybe the exponent in the original problem is \(\frac{1}{3}\) instead of \(\frac{1}{2}\). Let's assume the problem is \((x^{2}y^{3})^{\frac{1}{3}}\)
Then \((x^{2}y^{3})^{\frac{1}{3}}=x^{\frac{2}{3}}y^{1}=y\sqrt[3]{x^{2}}\) (since \(x^{\frac{2}{3}}=\sqrt[3]{x^{2}}\))
And since \(y^{3}\) is under the cube - root (if the exponent is \(\frac{1}{3}\)), \(y\) can be any real number, but in the original problem, if the exponent is \(\frac{1}{2}\), \(y\) must be non - negative. But the option A is \(y\sqrt[3]{x^{2}}\), maybe the exponent is \(\frac{1}{3}\). So if we assume the problem is \((x^{2}y^{3})^{\frac{1}{3}}\)

Step 1: Apply the power of a power rule \((a^m)^n=a^{mn}\)

\((x^{2}y^{3})^{\frac{1}{3}}=x^{\frac{2}{3}}y^{1}\)

Step 2: Rewrite \(x^{\frac{2}{3}}\) as a cube - root

\(x^{\frac{2}{3}}=\sqrt[3]{x^{2}}\), so \(x^{\frac{2}{3}}y^{1}=y\sqrt[3]{x^{2}}\)

Answer:

D. 343

Question 2