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Math  /  Algebra

QuestionSimplify. 5y48y3−27y55 y \sqrt{48 y^{3}}-\sqrt{27 y^{5}}
Assume that the variable represents a positive real number.

Studdy Solution

STEP 1

1. The variable y y represents a positive real number.
2. We need to simplify the expression by simplifying the square roots and combining like terms if possible.

STEP 2

1. Simplify the square root expressions individually.
2. Combine the simplified expressions.

STEP 3

Simplify the first square root expression 48y3 \sqrt{48 y^3} .
First, factor 48 48 into its prime factors:
48=24×3 48 = 2^4 \times 3
Rewrite the expression under the square root:
48y3=24×3×y3 \sqrt{48 y^3} = \sqrt{2^4 \times 3 \times y^3}
Separate the perfect squares from the non-perfect squares:
=(22)2×3×y2×y = \sqrt{(2^2)^2 \times 3 \times y^2 \times y}
Take the square root of the perfect squares:
=22×y×3y = 2^2 \times y \times \sqrt{3y}
=4y3y = 4y \sqrt{3y}

STEP 4

Simplify the second square root expression 27y5 \sqrt{27 y^5} .
First, factor 27 27 into its prime factors:
27=33 27 = 3^3
Rewrite the expression under the square root:
27y5=33×y5 \sqrt{27 y^5} = \sqrt{3^3 \times y^5}
Separate the perfect squares from the non-perfect squares:
=32×3×(y2)2×y = \sqrt{3^2 \times 3 \times (y^2)^2 \times y}
Take the square root of the perfect squares:
=3×y2×3y = 3 \times y^2 \times \sqrt{3y}
=3y23y = 3y^2 \sqrt{3y}

STEP 5

Substitute the simplified square root expressions back into the original expression:
5y×4y3y−3y23y 5y \times 4y \sqrt{3y} - 3y^2 \sqrt{3y}
=20y23y−3y23y = 20y^2 \sqrt{3y} - 3y^2 \sqrt{3y}

STEP 6

Combine like terms:
=(20y2−3y2)3y = (20y^2 - 3y^2) \sqrt{3y}
=17y23y = 17y^2 \sqrt{3y}
The simplified expression is:
17y23y \boxed{17y^2 \sqrt{3y}}

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