Question

can someone tell me if I have solved this correctly?

Simplify the following expression...
sqrt(-6) * sqrt(-42) = (-126)

Answer 1

\[\sqrt{?}\sqrt{-6}*\sqrt{42}= \sqrt{-6}=\sqrt{-1}*\sqrt{6}= \sqrt{-1}* \sqrt{2}*\sqrt{3} =i *\sqrt{2} *\sqrt{3} = 3i \sqrt \sqrt{-42}=\sqrt{-1}\sqrt{42} = \sqrt{-1} *\sqrt{2} * \sqrt{21} = i *\sqrt{2} *\sqrt{21} = 21i \sqrt{-6}*\sqrt{-42} = 3i \sqrt{2} * 21i \sqrt{2} = (21*3) (\sqrt{2} \sqrt{2} )(i*i)= 63*2*-1 =126 *-1= -126\]

Answer 2

0r would the answer be 1/7y^2 * 21x^3 * 1/y3 =1/7y^2 *1y^3 = 1/7y^5 *21x^ 3

Answer 3

\[\sqrt{-6} \times \sqrt{-42} \implies \sqrt{(-6) \times (-42)} \implies \sqrt{6 \times 42} \implies \sqrt{ 6 \times 6 \times 7 = ??}\]

Answer 4

\[\sqrt{\underline{6 \times 6} \times 7} = ??\]

Answer 5

252?

Answer 6

You can pull out 6 there..

In square root out of two pairs you can pull out one out of the square root brackets like:

\[\sqrt{\underline{x \times x} \times y} \implies x \sqrt{y}\]

Answer 7

\[6\sqrt{6*7}\] ?

Answer 8

Once you pulled out 6 you will not write 6 in the square root brackets...

Answer 9

\[6\sqrt{7}\] ?

Answer 10

In square root brackets after pulling one from the pair we delete the pair in the square root brackets:

\[\sqrt{4 \times 4 \times 5} \implies 4 \sqrt{5}\]

Answer 11

Yes that is right..

Answer 12

thank you:)

Answer 13

Welcome dear..