How do you simplify: $\sqrt{12} + 2 \sqrt{27} - 5 \sqrt{48}$ $sqrt12+2sqrt27-5sqrt48$ ?

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Answer 1 · 2017-04-06T05:59:08

$- 12 \sqrt{3}$ $-12sqrt3$

Let A $= \sqrt{12} + 2 \sqrt{27} - 5 \sqrt{48}$ $=sqrt12+2sqrt27-5sqrt48$

When simplifying roots of integers it if often helpful to express the integer in terms of its prime factors.

$\sqrt{12} = \sqrt{2 \times 2 \times 3} = 2 \sqrt{3}$ $sqrt12 = sqrt(2xx2xx3) = 2sqrt3$

$\sqrt{27} = \sqrt{3 \times 3 \times 3} = 3 \sqrt{3}$ $sqrt27= sqrt(3xx3xx3) = 3sqrt3$

$\sqrt{48} = \sqrt{2 \times 2 \times 2 \times 2 \times 3} = 2 \times 2 \sqrt{3}$ $sqrt48 = sqrt(2xx2xx2xx2xx3) = 2xx2sqrt3$

Hence A $= 2 \sqrt{3} + 2 \times 3 \sqrt{3} - 5 \times 4 \sqrt{3}$ $= 2sqrt3 + 2xx3sqrt3 - 5xx4sqrt3$

$= 2 \sqrt{3} + 6 \sqrt{3} - 20 \sqrt{3}$ $= 2sqrt3+6sqrt3-20sqrt3$

$= - 12 \sqrt{3}$ $=-12sqrt3$

How do you simplify: √12+2√27−5√48sqrt12+2sqrt27-5sqrt48 ?