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maw [93]
3 years ago
10

What is the following simplified product? Assume x greater-than-or-equal-to 0 (StartRoot 10 x Superscript 4 Baseline EndRoot min

us x StarRoot 5 x squared EndRoot) (2 StartRoot 15 x Superscript 4 Baseline EndRoot + StartRoot 3 x cubed EndRoot) 10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot + x squared StartRoot 15 x EndRoot 11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus x Superscript 4 Baseline StartRoot 75 EndRoot + x squared StartRoot 15 EndRoot 10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x squared StartRoot 15 EndRoot 11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x cubed StartRoot 15 x EndRoot
Mathematics
1 answer:
9966 [12]3 years ago
8 0

Answer:

\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

Step-by-step explanation:

To find:

Simplified product of:

(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})

Solution:

First of all, let us have a look at some of the formula:

1. (a+b) (c+d) = ac+bc+ad+bd

2. a^b\times a^c =a^{b+c }

3. \sqrt{a^{2b}} = \sqrt{a^b.a^b}=a^b

4. \sqrt a  \times \sqrt b = \sqrt{a\times b}

Now, let us apply the above formula to solve the given expression.

(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\\Rightarrow(\sqrt{10x^4})(2\sqrt{15x^4})+(\sqrt{10x^4})(\sqrt{3x^3})-(x\sqrt{5x^2})(2\sqrt{15x^4})-(x\sqrt{5x^2})(\sqrt{3x^3})\\\\\Rightarrow2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\\Rightarrow\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

The answer is:

\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

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