simplify the radical expression. 5. √250h^4k^5 6. √14q x 2√4q 7. √63x^15y^9/7xy^11 8. √3+4√3 5√6 5√3 3√6 3√3 9. 2√6+3√96 14√6 14√96 5√96 50√6 10. (8 +√11 )(8 –√11 ) 53 75 + 16√11 –57 64 + √11
For (5), is this the question? \[\sqrt{250h^4k^5}\]
yea
Simple foil out question 11, remembering that: \[\sqrt{11}\times \sqrt{11}=11\]
\[\sqrt{250h^4k^5} = \sqrt{25 \times 10 (h^2)^2(k^2)^2k} = \sqrt{(5h^2k^2)^2\times 10k}\]\[ =\sqrt{(5h^2k^2)^2}\times \sqrt{10k} =? \]
You should remember that\[\sqrt{a} \times \sqrt{b}=\sqrt{ab}\] and \[\sqrt{a} \times \sqrt{a}=a\]
25hk√10k
Nope ._. \[\sqrt{a^2} =a\] Can you try again?
5√10h^4k^5
Nope.... try a =5h^2k^2 what do you get?
hk√125
Incorrect again.... \[\sqrt{(5h^2k^2)^2} \times \sqrt{10k} = 5h^2k^2 \sqrt{10k}\] Since 5h^2k^2 are ALL squared... Got it?
so did i get it right
Nope ._.
5h^2k^2√10k
Yes. this is right
thanks
For (6) is this the question? \[\sqrt{14q} \times 2\sqrt{4q}\]
yea
I have 5 and 6.
\[\sqrt{14q} \times 2\sqrt{4q}=\sqrt{14q} \times 2\sqrt{2^2q}=\sqrt{14q} \times 4\sqrt{q}\]\[=4\sqrt{(14q)\times q} = 4\sqrt{14q^2} = ?\]
can you the answers
give the
Can you solve it yourself first?
I didn't mean I didn't want to help you.. but I think you should at least try to do it once first...
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