How do you Multiply and simplify the product [2sqrt(7) – sqrt(3)]^2?
Can someone please tell me how to do this problem.
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haha, my answer wouldn’t help u i think, it’s ’16hours ago’ ^^
but whatever, i still get the 2 points =P
differ with previous answer, i don’t use form, but form would b faster,,
[2sqrt(7) – sqrt(3)]^2
= [2sqrt(7) – sqrt(3)] * [2sqrt(7) – sqrt(3)]
= [2sqrt(7) + (-sqrt(3))] * [2sqrt(7) + (-sqrt(3))]
__multiply them
= 2sqrt(7)*2sqrt(7) + 2sqrt(7)*(-sqrt(3)) + (-sqrt(3))*2sqrt(7) + (-sqrt(3))*(-sqrt(3))
= 2sqrt(7)*2sqrt(7) + 2*2sqrt(7)*(-sqrt(3)) + (-sqrt(3))*(-sqrt(3))
= 4*7 + (-4sqrt(21)) + 3
= 28 – 4sqrt(21) + 3
= 31 – 4sqrt(21)
…done!
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You are squaring a binomial, so you must think of your answer in this form:
(a – b)^2 = a^2 – 2ab + b^2
So your answer is:
[2sqrt(7)]^2 – 2*[2sqrt(7)]*[sqrt(3)] + [sqrt(3)]^2
Which simplifies to:
4*7 – 4sqrt(21) + 3
=28 – 4sqrt(21) + 3
(final answer) = 31 – 4sqrt(21)
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