3. $$ f(x)=0 $$ has the following roots, $$ x=\frac{-6 \pm \sqrt{1-4 k^{2}}}{4} $$, for which values of $$ k $$ will the roots be equal? UESTION 2 1. Solve the following without the use of a calculator. 2.1.1. $$ \frac{\sqrt{75} imes \sqrt{49}}{\sqrt{147}} $$ 2.1.2. $$ \frac{5^{2 x+2}-25^{x}}{4.5^{2 x}} $$

Question

3. $$ f(x)=0 $$ has the following roots, $$ x=\frac{-6 \pm \sqrt{1-4 k^{2}}}{4} $$, for which values of $$ k $$ will the roots be equal? UESTION 2 1. Solve the following without the use of a calculator. 2.1.1. $$ \frac{\sqrt{75} 	imes \sqrt{49}}{\sqrt{147}} $$ 2.1.2. $$ \frac{5^{2 x+2}-25^{x}}{4.5^{2 x}} $$

UpStudy AI · Accepted Answer

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 $$ \frac{5^{2x+2} - 25^{x}}{4 \cdot 5^{2x}} $$: $$ 6 $$
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 $$ 6 $$.

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