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Mathematics 14 Online
OpenStudy (anonymous):

HELp offering medals. if question is correct. Simplify the radical expression.

OpenStudy (anonymous):

OpenStudy (anonymous):

answer choices are: and d is not the correct answer so...

OpenStudy (anonymous):

a

OpenStudy (anonymous):

b

OpenStudy (anonymous):

c

OpenStudy (anonymous):

im not going to give you d because its wrong.

OpenStudy (anonymous):

Look : Just do this work : we have : 16q under Radical * 14q un R \[\sqrt{16q}*\sqrt{14q}=\sqrt{16*14}

OpenStudy (austinl):

you leave the 2 alone, so we will have. \(2\sqrt{\underline{numbers}}\) Now what is \(14q\times4q\)?

OpenStudy (anonymous):

\[\sqrt{16q}*\sqrt{14q}= ?\]

OpenStudy (anonymous):

have : \[\sqrt{56q^2}=q \sqrt{56}=\sqrt2q{14}\]=d

OpenStudy (anonymous):

so whats the answer c?

OpenStudy (austinl):

What is \(14q\times4q\)? That will be under the square root, and that is multiplied by 2.

OpenStudy (anonymous):

Yea !!! yea !

OpenStudy (anonymous):

@Kenziehulse :Your answer is right !:)

OpenStudy (austinl):

It will give, \(2\sqrt{56q^2}\) which can be rewritten as \(2\times\sqrt{56}\times\sqrt{q^2}\) \(\sqrt{56}=2\sqrt{14}\) \(\sqrt{q^2}=q\) So now we have, \(2\times2\sqrt{14}\times q=~?\)

OpenStudy (anonymous):

k im closin gbut im opening another still offering medals for every one of my questions .:)

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