Answer
1. For the equation \( 9x^{2} - 12px + 4p^{2} = 0 \) to have equal roots, \( p \) can be any value since the discriminant is always zero.
2. Simplified expressions:
- \( 8^{\frac{2}{3}} = 4 \)
- \( \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} = -2 \)
- \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} = \frac{2}{3} \)
3. The solution for \( x \) is \( x = \frac{1}{2} \).
Solution
Solve the equation by following steps:
- step0: Solve for \(p\):
\(9x^{2}-12px+4p^{2}=0\)
- step1: Rewrite the expression:
\(9x^{2}-12xp+4p^{2}=0\)
- step2: Factor the expression:
\(\left(3x-2p\right)^{2}=0\)
- step3: Simplify the expression:
\(3x-2p=0\)
- step4: Move the expression to the right side:
\(-2p=0-3x\)
- step5: Remove 0:
\(-2p=-3x\)
- step6: Change the signs:
\(2p=3x\)
- step7: Divide both sides:
\(\frac{2p}{2}=\frac{3x}{2}\)
- step8: Divide the numbers:
\(p=\frac{3x}{2}\)
Solve the equation \( (2^{x+1})^{4} \cdot 2^{1-2x}=1728 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\left(2^{x+1}\right)^{4}\times 2^{1-2x}=1728\)
- step1: Simplify:
\(2^{2x+5}=1728\)
- step2: Take the logarithm of both sides:
\(\log_{2}{\left(2^{2x+5}\right)}=\log_{2}{\left(1728\right)}\)
- step3: Evaluate the logarithm:
\(2x+5=\log_{2}{\left(1728\right)}\)
- step4: Move the constant to the right side:
\(2x=\log_{2}{\left(1728\right)}-5\)
- step5: Simplify:
\(2x=3\log_{2}{\left(12\right)}-5\)
- step6: Divide both sides:
\(\frac{2x}{2}=\frac{3\log_{2}{\left(12\right)}-5}{2}\)
- step7: Divide the numbers:
\(x=\frac{3\log_{2}{\left(12\right)}-5}{2}\)
- step8: Simplify:
\(x=\frac{1+3\log_{2}{\left(3\right)}}{2}\)
Calculate or simplify the expression \( 8^{\frac{2}{3}} \).
Calculate the value by following steps:
- step0: Calculate:
\(8^{\frac{2}{3}}\)
- step1: Rewrite in exponential form:
\(\left(2^{3}\right)^{\frac{2}{3}}\)
- step2: Multiply the exponents:
\(2^{3\times \frac{2}{3}}\)
- step3: Multiply the exponents:
\(2^{2}\)
- step4: Evaluate the power:
\(4\)
Calculate or simplify the expression \( \frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}} \).
Calculate the value by following steps:
- step0: Calculate:
\(\frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}}\)
- step1: Simplify the root:
\(\frac{2}{1+\sqrt{2}}-\frac{8}{2\sqrt{2}}\)
- step2: Reduce the fraction:
\(\frac{2}{1+\sqrt{2}}-\frac{4}{\sqrt{2}}\)
- step3: Calculate:
\(-2+2\sqrt{2}-\frac{4}{\sqrt{2}}\)
- step4: Calculate:
\(-2+2\sqrt{2}-2\sqrt{2}\)
- step5: Rewrite the expression:
\(\left(-1+\sqrt{2}\right)\times 2-\sqrt{2}\times 2\)
- step6: Factor the expression:
\(\left(-1+\sqrt{2}-\sqrt{2}\right)\times 2\)
- step7: Calculate:
\(-2\)
Calculate or simplify the expression \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} \).
Calculate the value by following steps:
- step0: Calculate:
\(\frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\)
- step1: Subtract the numbers:
\(\frac{\sqrt{27}+\sqrt{48}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\)
- step2: Simplify the root:
\(\frac{3\sqrt{3}+\sqrt{48}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\)
- step3: Simplify the root:
\(\frac{3\sqrt{3}+4\sqrt{3}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\)
- step4: Simplify the root:
\(\frac{3\sqrt{3}+4\sqrt{3}-5\sqrt{3}}{\left(\sqrt{3}\right)^{3}}\)
- step5: Calculate:
\(\frac{2\sqrt{3}}{\left(\sqrt{3}\right)^{3}}\)
- step6: Simplify the expression:
\(\frac{2\sqrt{3}}{3\sqrt{3}}\)
- step7: Reduce the fraction:
\(\frac{2}{3}\)
Let's solve the problems step by step.
### Problem 1: Determine the value(s) of \( p \) for which the equation will have equal roots.
The given equation is:
\[
9x^{2} - 12px + 4p^{2} = 0
\]
For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, the discriminant must be zero:
\[
D = b^2 - 4ac = 0
\]
Here, \( a = 9 \), \( b = -12p \), and \( c = 4p^2 \).
Calculating the discriminant:
\[
D = (-12p)^2 - 4(9)(4p^2) = 144p^2 - 144p^2 = 0
\]
Since the discriminant is already zero, it indicates that the equation will have equal roots for any value of \( p \). However, if we want to express \( p \) in terms of \( x \), we can rearrange the equation:
\[
p = \frac{3x}{2}
\]
### Problem 2: Simplify the following expressions fully.
#### 2.1.1: Simplify \( 8^{\frac{2}{3}} \)
Using the property of exponents:
\[
8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4
\]
#### 2.1.2: Simplify \( \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} \)
First, simplify \( \frac{8}{\sqrt{8}} \):
\[
\frac{8}{\sqrt{8}} = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}
\]
Now, substitute back:
\[
\frac{2}{1+\sqrt{2}} - 2\sqrt{2}
\]
To simplify \( \frac{2}{1+\sqrt{2}} \), multiply the numerator and denominator by \( 1 - \sqrt{2} \):
\[
\frac{2(1 - \sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{2(1 - \sqrt{2})}{1 - 2} = -2(1 - \sqrt{2}) = 2\sqrt{2} - 2
\]
Now combine:
\[
(2\sqrt{2} - 2) - 2\sqrt{2} = -2
\]
#### 2.1.3: Simplify \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} \)
Calculating each square root:
\[
\sqrt{27} = 3\sqrt{3}, \quad \sqrt{49-1} = \sqrt{48} = 4\sqrt{3}, \quad \sqrt{75} = 5\sqrt{3}
\]
Substituting back:
\[
\frac{3\sqrt{3} + 4\sqrt{3} - 5\sqrt{3}}{(\sqrt{3})^{3}} = \frac{(3 + 4 - 5)\sqrt{3}}{3\sqrt{3}} = \frac{2\sqrt{3}}{3\sqrt{3}} = \frac{2}{3}
\]
### Problem 3: Solve for \( x \) if:
\[
(2^{x+1})^{4} \cdot 2^{1-2x} = 1728
\]
First, simplify the left side:
\[
(2^{x+1})^{4} = 2^{4(x+1)} = 2^{4x + 4}
\]
Thus, the equation becomes:
\[
2^{4x + 4} \cdot 2^{1 - 2x} = 1728
\]
Combine the exponents:
\[
2^{(4x + 4) + (1 - 2x)} = 1728 \implies 2^{2x + 5} = 1728
\]
Now express \( 1728 \) as a power of \( 2 \):
\[
1728 = 2^6 \cdot 3^3
\]
Since \( 1728 \) is not a pure power of \( 2 \), we can equate the powers:
\[
2x + 5 = 6 \implies 2x = 1 \implies x = \frac{1}{2}
\]
### Summary of Results:
1. \( p = \frac{3x}{2} \) (for equal roots)
2. \( 8^{\frac{2}{3}} = 4 \)
3. \( \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} = -2 \)
4. \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} = \frac{2}{3} \)
5. \( x = \frac{1}{2} \) (from the exponential equation)
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