Q. \(2:x = x:50\). What one number can replace \(x\)?

Q: What value(s) of x satisfy \frac{2}{x} = \frac{x}{50} ?

Solve \frac{2}{x}=\frac{x}{50} . Cross-multiply: 2\cdot50 = x^2, so x^2=100. Thus x=10 or x=-10. If only one (positive) number is wanted, choose 10.

Q: Why can't x=0 be a solution?

x cannot be 0 because the equation has x in a denominator. Division by zero is undefined, so x=0 is excluded from the domain.

Q: How do you use cross-multiplication to solve \frac{2}{x}=\frac{x}{50} ?

Cross-multiply to get 2 \cdot 50 = x \cdot x, i.e., 100 = x^2. Then take square roots: x = \pm\sqrt{100} = \pm 10.

Q: Are both +10 and -10 valid solutions?

Yes. Plugging in x=10 gives 2/10=10/50=0.2. Plugging in x=-10 gives 2/(-10)=(-10)/50=-0.2. Both satisfy the equation.

Q: If the problem asks for "one number" to replace x, which should you give?

Usually give the positive root x=10 unless negatives are allowed or specifically requested. Many contexts mean the principal (positive) solution.

Q: How is x related to 2 and 50 in terms of means?

How is x related to 2 and 50 in terms of means?

Q: How can you quickly check your answers?

Substitute back into \frac{2}{x}=\frac{x}{50} . For x=10: 2/10=10/50. For x=-10: 2/(-10)=(-10)/50. Both sides match, so the answers are correct.

Q: Could there be complex solutions to \frac{2}{x}=\frac{x}{50} ?

From x^2=100, the only solutions are x=\pm10, which are real. No additional complex solutions arise because the polynomial is degree 2 with real roots ±10.

Answer

Convert to fractions: \( \frac{2}{x} = \frac{x}{50} \). Cross-multiply: \(2\cdot 50 = x^2\), so \(x^2 = 100\). Thus \(x = \pm 10\). One number that can replace x is 10.

Detailed Explanation

Step-by-step solution

Write the proportion as an equation of fractions:
\[
\frac{2}{x} = \frac{x}{50}
\]
Use cross-multiplication (multiply both sides by x and by 50) to eliminate denominators and obtain a polynomial equation. Multiplying both sides by \(x\cdot 50\) gives:
\[
2 \cdot 50 = x \cdot x
\]
which simplifies to
\[
100 = x^{2}
\]
Solve the quadratic equation by taking square roots of both sides. The equation \(x^{2} = 100\) yields two real solutions:
\[
x = \sqrt{100} \quad \text{or} \quad x = -\sqrt{100}
\]
so
\[
x = 10 \quad \text{or} \quad x = -10
\]
Interpret the solutions: both \(x = 10\) and \(x = -10\) satisfy the algebraic equation. If the context requires a positive quantity (for example, a length or a positive ratio), choose the positive solution \(x = 10\). Otherwise both values are valid solutions to the given proportion.

Answer: x = 10 or x = -10. The one positive number that can replace x is 10.

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FAQs

What value(s) of x satisfy \( \frac{2}{x} = \frac{x}{50} \)?

Solve \( \frac{2}{x}=\frac{x}{50} \). Cross-multiply: \(2\cdot50 = x^2\), so \(x^2=100\). Thus \(x=10\) or \(x=-10\). If only one (positive) number is wanted, choose 10.

Why can't \(x=0\) be a solution?

\(x\) cannot be 0 because the equation has \(x\) in a denominator. Division by zero is undefined, so \(x=0\) is excluded from the domain.

How do you use cross-multiplication to solve \( \frac{2}{x}=\frac{x}{50} \)?

Cross-multiply to get \(2 \cdot 50 = x \cdot x\), i.e., \(100 = x^2\). Then take square roots: \(x = \pm\sqrt{100} = \pm 10\).

Are both \(+10\) and \(-10\) valid solutions?

Yes. Plugging in \(x=10\) gives \(2/10=10/50=0.2\). Plugging in \(x=-10\) gives \(2/(-10)=(-10)/50=-0.2\). Both satisfy the equation.

If the problem asks for "one number" to replace \(x\), which should you give?

Usually give the positive root \(x=10\) unless negatives are allowed or specifically requested. Many contexts mean the principal (positive) solution.

How is \(x\) related to 2 and 50 in terms of means?

How can you quickly check your answers?

Substitute back into \( \frac{2}{x}=\frac{x}{50} \). For \(x=10\): \(2/10=10/50\). For \(x=-10\): \(2/(-10)=(-10)/50\). Both sides match, so the answers are correct.

Could there be complex solutions to \( \frac{2}{x}=\frac{x}{50} \)?

From \(x^2=100\), the only solutions are \(x=\pm10\), which are real. No additional complex solutions arise because the polynomial is degree 2 with real roots ±10.

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