Answer
For the equation \( f(x) = 0 \), the roots are equal when \( k = -\frac{1}{2} \) or \( k = \frac{1}{2} \). The simplified expressions are \( 5 \) and \( 6 \).
Solution
Calculate the value by following steps:
- step0: Calculate:
\(\frac{\sqrt{75}\times \sqrt{49}}{\sqrt{147}}\)
- step1: Simplify the root:
\(\frac{5\sqrt{3}\times \sqrt{49}}{\sqrt{147}}\)
- step2: Simplify the root:
\(\frac{5\sqrt{3}\times 7}{\sqrt{147}}\)
- step3: Simplify the root:
\(\frac{5\sqrt{3}\times 7}{7\sqrt{3}}\)
- step4: Multiply the terms:
\(\frac{35\sqrt{3}}{7\sqrt{3}}\)
- step5: Reduce the fraction:
\(5\)
Calculate or simplify the expression \( \frac{5^{2x+2}-25^{x}}{4 \cdot 5^{2x}} \).
Simplify the expression by following steps:
- step0: Solution:
\(\frac{5^{2x+2}-25^{x}}{4\times 5^{2x}}\)
- step1: Subtract the terms:
\(\frac{24\times 5^{2x}}{4\times 5^{2x}}\)
- step2: Reduce the fraction:
\(6\)
Solve the equation \( 1-4k^{2}=0 \).
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
\(1-4k^{2}=0\)
- step1: Add or subtract both sides:
\(-4k^{2}=-1\)
- step2: Divide both sides:
\(\frac{-4k^{2}}{-4}=\frac{-1}{-4}\)
- step3: Divide the numbers:
\(k^{2}=\frac{1}{4}\)
- step4: Simplify the expression:
\(k=\pm \sqrt{\frac{1}{4}}\)
- step5: Simplify the expression:
\(k=\pm \frac{1}{2}\)
- step6: Separate into possible cases:
\(\begin{align}&k=\frac{1}{2}\\&k=-\frac{1}{2}\end{align}\)
- step7: Rewrite:
\(k_{1}=-\frac{1}{2},k_{2}=\frac{1}{2}\)
Let's solve the problems step by step.
### Problem 1: Roots of the Equation
The roots of the equation \( f(x) = 0 \) are given by:
\[
x = \frac{-6 \pm \sqrt{1 - 4k^2}}{4}
\]
To find the values of \( k \) for which the roots are equal, we need the expression under the square root (the discriminant) to be zero:
\[
1 - 4k^2 = 0
\]
Now, let's solve for \( k \):
1. Rearranging the equation gives:
\[
4k^2 = 1
\]
2. Dividing both sides by 4:
\[
k^2 = \frac{1}{4}
\]
3. Taking the square root of both sides:
\[
k = \pm \frac{1}{2}
\]
Thus, the values of \( k \) for which the roots are equal are:
\[
k = -\frac{1}{2} \quad \text{and} \quad k = \frac{1}{2}
\]
### Problem 2: Simplifying Expressions
#### 2.1.1: Simplifying \( \frac{\sqrt{75} \times \sqrt{49}}{\sqrt{147}} \)
Let's simplify this step by step:
1. Calculate \( \sqrt{75} \):
\[
\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}
\]
2. Calculate \( \sqrt{49} \):
\[
\sqrt{49} = 7
\]
3. Calculate \( \sqrt{147} \):
\[
\sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}
\]
4. Now substitute these values into the expression:
\[
\frac{\sqrt{75} \times \sqrt{49}}{\sqrt{147}} = \frac{(5\sqrt{3}) \times 7}{7\sqrt{3}} = \frac{35\sqrt{3}}{7\sqrt{3}} = 5
\]
So, the simplified result is:
\[
5
\]
#### 2.1.2: Simplifying \( \frac{5^{2x+2} - 25^{x}}{4 \cdot 5^{2x}} \)
1. Recognize that \( 25^x = (5^2)^x = 5^{2x} \):
\[
25^x = 5^{2x}
\]
2. Substitute this into the expression:
\[
\frac{5^{2x+2} - 5^{2x}}{4 \cdot 5^{2x}} = \frac{5^{2x}(5^2 - 1)}{4 \cdot 5^{2x}} = \frac{5^{2x} \cdot 24}{4 \cdot 5^{2x}} = \frac{24}{4} = 6
\]
So, the simplified result is:
\[
6
\]
### Summary of Results
1. Values of \( k \): \( k = -\frac{1}{2}, \frac{1}{2} \)
2. Simplified expression \( \frac{\sqrt{75} \times \sqrt{49}}{\sqrt{147}} \): \( 5 \)
3. Simplified expression \( \frac{5^{2x+2} - 25^{x}}{4 \cdot 5^{2x}} \): \( 6 \)
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