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You are watching: Square root of 108 in radical form

Let's take u221a(108) with these actions. First, we recompose 108 as a product of a perfect square and one more number. Notice that 108 is divisible by 4, and 4 is a perfect square, so we deserve to recompose 108 as 4 u00d7 27.u221a(108) = u221a(4 u00d7 27)Now, create the square root as a product of square roots using our rule.u221a(4 u00d7 27) = u221a(4) u00d7 u221a(27)Next, we evaluate u221a(4) as 2.u221a(4) u00d7 u221a(27) = 2 u00d7 u221a(27)Now, we start over for u221a(27). We have the right to recompose 27 as 9 u00d7 3, where 9 is a perfect square, then take it with our measures.u221a(108) = 2 u00d7 u221a(27) = 2 u00d7 u221a(9 u00d7 3) = 2 u00d7 u221a(9) u00d7 u221a(3) = 2 u00d7 3 u00d7 u221a(3) = 6 u221a(3)tysm..#gozmit"},"id":4034319,"content":"Hey mate! Here's your answer!! Here are some steps. First, we recompose 108 as a product of a perfect square and another number. Notice that 108 is divisible by 4, and 4 is a perfect square, so we have the right to recreate 108 as 4 u00d7 27.u221a(108) = u221a(4 u00d7 27)Now, write the square root as a product of square roots using our dominion.u221a(4 u00d7 27) = u221a(4) u00d7 u221a(27)Next, we evaluate u221a(4) as 2.u221a(4) u00d7 u221a(27) = 2 u00d7 u221a(27)Now, we start over for u221a(27). We have the right to recreate 27 as 9 u00d7 3, wbelow 9 is a perfect square, then take it via our steps.u221a(108) = 2 u00d7 u221a(27) = 2 u00d7 u221a(9 u00d7 3) = 2 u00d7 u221a(9) u00d7 u221a(3) = 2 u00d7 3 u00d7 u221a(3) = 6 u00d7 u221a(3)Due to the fact that 3 doesn't have any kind of perfect square factors, u221a(3) is as simplified as feasible, so we have that the square root of 108 in radical create is 6 u00d7 u221a(3), or 6u221a(3).hope \: it \: helps \: you.u270c u270c #BE ptcouncil.net">" data-test="answer-box-list"> heya..
Let"s take √(108) through these measures. First, we recompose 108 as a product of a perfect square and also an additional number. Notice that 108 is divisible by 4, and 4 is a perfect square, so we deserve to recompose 108 as 4 × 27.

√(108) = √(4 × 27)

Now, create the square root as a product of square roots utilizing our dominance.

√(4 × 27) = √(4) × √(27)

Next off, we evaluate √(4) as 2.

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√(4) × √(27) = 2 × √(27)

Now, we start over for √(27). We have the right to rewrite 27 as 9 × 3, wright here 9 is a perfect square, then take it through our procedures.

√(108) = 2 × √(27) = 2 × √(9 × 3) = 2 × √(9) × √(3) = 2 × 3 × √(3) = 6 √(3)tysm..#gozmit