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Sagot :
To solve the given problem, where we are asked to find the product of [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex] with the assumption that [tex]\(b \geq 0\)[/tex], we can proceed as follows:
1. Understand the expression:
- We need to compute the product of two square roots: [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex].
2. Apply the property of square roots:
- Recall the property: [tex]\(\sqrt{a} \cdot \sqrt{a} = a\)[/tex].
- This property tells us that the product of two identical square roots results in the number under the square root.
3. Apply this property to our problem:
- Given two identical square roots: [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex].
- According to the property mentioned above, [tex]\(\sqrt{b} \cdot \sqrt{b} = b\)[/tex].
4. State the result:
- Therefore, the product [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex] simplifies to [tex]\(b\)[/tex].
Now, we can compare this to the given options:
- [tex]\(b \sqrt{b}\)[/tex]
- [tex]\(2 \sqrt{b}\)[/tex]
- [tex]\(b\)[/tex]
- [tex]\(b^2\)[/tex]
Among the given choices, the correct answer is:
[tex]\[ b \][/tex]
Hence, the product [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex] is indeed [tex]\(b\)[/tex].
1. Understand the expression:
- We need to compute the product of two square roots: [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex].
2. Apply the property of square roots:
- Recall the property: [tex]\(\sqrt{a} \cdot \sqrt{a} = a\)[/tex].
- This property tells us that the product of two identical square roots results in the number under the square root.
3. Apply this property to our problem:
- Given two identical square roots: [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex].
- According to the property mentioned above, [tex]\(\sqrt{b} \cdot \sqrt{b} = b\)[/tex].
4. State the result:
- Therefore, the product [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex] simplifies to [tex]\(b\)[/tex].
Now, we can compare this to the given options:
- [tex]\(b \sqrt{b}\)[/tex]
- [tex]\(2 \sqrt{b}\)[/tex]
- [tex]\(b\)[/tex]
- [tex]\(b^2\)[/tex]
Among the given choices, the correct answer is:
[tex]\[ b \][/tex]
Hence, the product [tex]\(\sqrt{b} \cdot \sqrt{b}\)[/tex] is indeed [tex]\(b\)[/tex].
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