Remove all perfect squares from inside the square root. Assume b is positive. 80 2 = 80b 2 =square root of, 80, b, squared, end square root, equals
√80 = 4√5
√(b^2) = b
√80b^2 = b√80
To remove perfect squares from inside the square root, we can factorize the number under the square root and simplify it.
Let's start with √80:
√80 = √(16 × 5)
Since 16 is a perfect square, we can take its square root outside of the square root:
√80 = 4√5
Now, let's consider (√b)²:
(√b)² = b
So, combining these results, we get:
√(80b²) = √80 × (√b)² = 4√5b
To remove perfect squares from inside the square root, you need to express the number inside the square root as the product of a perfect square and another number.
Let's work through the example step by step:
√(80b^2)
First, we factor out all perfect squares from 80. The perfect square factors of 80 are 4 and 16.
√(4 * 16 * b^2)
Next, we can rewrite the square root as a product of separate square roots.
√(4) * √(16) * √(b^2)
Simplifying further:
2 * 4 * b
8b
Therefore, the square root of 80b^2 is equal to 8b.