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# what is the domain of the formula "square root of a * square root of b = square root of ab"?

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Guest Dec 9, 2014

#7
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If a and b are both negative then

$$\sqrt a\times \sqrt b=\sqrt{-|a|}\times \sqrt{-|b|}=i^2\sqrt{|a|\times |b|}=-\sqrt{|a| \times |b|}$$

But

$$\sqrt{a\times b}=\sqrt{-|a|\times -|b|}=\sqrt{|a|\times|b|}$$

i.e. the two are different, so not true when both are negative.

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Alan  Dec 9, 2014
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Been a while since I've toyed around with imaginary numbers, but I'm pretty sure that, staying on the real side of numbers and away from imaginary numbers, the domain of sqrt(a) * sqrt(b) equals [a > 0; b > 0], or all positive numbers for both 'a' and 'b'.

GoldenLeaf  Dec 9, 2014
#2
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a and b can be 0 tho

Guest Dec 9, 2014
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Yes Anon, a and b can equal 0. I blame my computer not having the 'greater than or equal to' keys >.<

But no numbers below 0, basically no negative numbers can equal 'a' OR 'b'.

GoldenLeaf  Dec 9, 2014
#4
+5

a and b can be negative number

x^2+1=0

x^2=-1

x = positive or negative square root of -1

In the 1500s .,an italian mathematician named rafael bombelli invented the imaginary number , which is now called i.  Square root = i implie that i^2=-1 after his invention , it became possible to find soloution for x^2+1=0: they are i and -i. The value of square root of -16 =square root of 16 *(-1)=square root of 16*i^2=4i

bUt the domain of a and b in that formula  is a>or=0 and b>or =0

Guest Dec 9, 2014
#5
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no one anwser my original anwser yet! I need more help!

Guest Dec 9, 2014
#6
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$$\sqrt{a}+\sqrt{b}=\sqrt{ab}$$

There is no domain here - the word does not make sense here - not to me anyway.

If you are only considering real numbers then a and be will be restricted as follows

$$a\ge 0 \qquad and \qquad b \ge 0$$

Melody  Dec 9, 2014
#7
+27042
+13

If a and b are both negative then

$$\sqrt a\times \sqrt b=\sqrt{-|a|}\times \sqrt{-|b|}=i^2\sqrt{|a|\times |b|}=-\sqrt{|a| \times |b|}$$

But

$$\sqrt{a\times b}=\sqrt{-|a|\times -|b|}=\sqrt{|a|\times|b|}$$

i.e. the two are different, so not true when both are negative.

.

Alan  Dec 9, 2014
#8
+5

Thank you guys!

Guest Dec 9, 2014