Question

?^2 * 3 + 5 = 34

Answer 1

its a number 1-7

Answer 2

To solve the equation \( x^2 \cdot 3 + 5 = 34 \), follow these steps:

1. Subtract 5 from both sides:
\[
x^2 \cdot 3 = 34 - 5
\]
\[
x^2 \cdot 3 = 29
\]

2. Divide both sides by 3:
\[
x^2 = \frac{29}{3}
\]

3. To find \( x \), take the square root of both sides:
\[
x = \pm \sqrt{\frac{29}{3}}
\]

4. You can simplify the square root:
\[
x = \pm \frac{\sqrt{29}}{\sqrt{3}} = \pm \frac{\sqrt{87}}{3}
\]

Thus, the solutions to the equation are:
\[
x = \frac{\sqrt{87}}{3} \quad \text{or} \quad x = -\frac{\sqrt{87}}{3}
\]

Answer 3

Given that the solution must be a number between 1 and 7, let's solve the equation while finding the value of \(x\) within that range.

Starting from the equation:
\[
3x^2 + 5 = 34
\]

1. Subtract 5 from both sides:
\[
3x^2 = 34 - 5
\]
\[
3x^2 = 29
\]

2. Divide both sides by 3:
\[
x^2 = \frac{29}{3}
\]

3. Take the square root:
\[
x = \pm \sqrt{\frac{29}{3}}
\]

Now, let's calculate \(x\) to see if it falls within the range of 1 to 7:
\[
x = \sqrt{\frac{29}{3}} \approx \sqrt{9.67} \approx 3.11
\]

Since \(\sqrt{\frac{29}{3}} \approx 3.11\), the positive root of the equation is indeed between 1 and 7.

Thus, the valid solution is:
\[
x \approx 3.11
\]