What is the simplest radical form of the square root of 128?
The Correct Answer and Explanation is:
The simplest radical form of the square root of 128 is: 128=82\sqrt{128} = 8\sqrt{2}128=82
Explanation
To simplify the square root of 128, we need to factor the number into its prime components and look for perfect squares. The goal is to find the largest perfect square factor of 128, then use the property: a⋅b=a⋅b\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}a⋅b=a⋅b
Step 1: Prime factorization of 128
We divide 128 by 2 repeatedly since it is even: 128÷2=6464÷2=3232÷2=1616÷2=88÷2=44÷2=22÷2=1128 \div 2 = 64 64 \div 2 = 32 32 \div 2 = 16 16 \div 2 = 8 8 \div 2 = 4 4 \div 2 = 2 2 \div 2 = 1128÷2=6464÷2=3232÷2=1616÷2=88÷2=44÷2=22÷2=1
So the prime factorization is: 128=2×2×2×2×2×2×2=27128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7128=2×2×2×2×2×2×2=27
Step 2: Group the factors
To simplify a square root, we look for pairs of factors since: a2=a\sqrt{a^2} = aa2=a
We can group the powers of 2: 27=(22)3⋅2=43⋅22^7 = (2^2)^3 \cdot 2 = 4^3 \cdot 227=(22)3⋅2=43⋅2
That means: 128=(22)3⋅2=43⋅2\sqrt{128} = \sqrt{(2^2)^3 \cdot 2} = \sqrt{4^3 \cdot 2}128=(22)3⋅2=43⋅2
Break it apart: 128=64⋅2=64⋅2=82\sqrt{128} = \sqrt{64 \cdot 2} = \sqrt{64} \cdot \sqrt{2} = 8\sqrt{2}128=64⋅2=64⋅2=82
Final Answer: 128=82\sqrt{128} = 8\sqrt{2}128=82
This is in simplest radical form because 2 is not a perfect square and cannot be simplified further. Simplifying radicals in this way helps express numbers in the most reduced and exact form, especially in algebra and geometry.
