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Take a factor out of the square root:
a) √6x^2, where x≥0
b)√9a^3
d)√50b^4
plz help 30 points will give brainliest

Sagot :

xKelvin

Answer:

Question A)

[tex]=\sqrt{6}x[/tex]

Question B)

[tex]=3a\sqrt{a}[/tex]

Question C)

[tex]=5\sqrt{2}b^2[/tex]

Step-by-step explanation:

A)

We are given:

[tex]\sqrt{6x^2}\, \text{ where } x\geq 0[/tex]

We can rewrite the expression:

[tex]=\sqrt{6}\cdot \sqrt{x^2}[/tex]

The square root and square will cancel each other out. Thus:

[tex]=\sqrt{6}x[/tex]

B)

We are given:

[tex]\sqrt{9a^3}[/tex]

Rewrite:

[tex]=\sqrt{9}\cdot \sqrt{a^3}[/tex]

Note that the square root of 9 is simply 3. We can also factor the second part:

[tex]=3\cdot \sqrt{a^2\cdot a}[/tex]

Rewriting:

[tex]=3\cdot\sqrt{a^2}\cdot\sqrt{a}[/tex]

Simplify:

[tex]=3a\sqrt{a}[/tex]

C)

We are given:

[tex]\sqrt{50b^4}[/tex]

Rewrite. Note that 50 = 25(2):

[tex]=\sqrt{25}\cdot \sqrt{2}\cdot \sqrt{b^4}[/tex]

Simplify. We can rewrite the factor as:

[tex]=5\cdot \sqrt{2}\cdot \sqrt{(b^2)^2}[/tex]

The square and square root will cancel out. Thus:

[tex]=5\sqrt{2}b^2[/tex]

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