Q. 181 - Intermediate Algebra

Question

In the following exercises, simplify.

$3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}$

Step-by-Step Solution

Answer

On simplification, we get $4y\sqrt{2}$

Step 1: Given information

Given the expression $3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}$

Step 2: Simplifying the radical and combine like terms

Simplifying radical and combining we get

$$\begin{gathered} 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=3\sqrt{64y^2·2}+4y\sqrt{81·2}-8\sqrt{49y^2·2} \\ 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=3·\sqrt{64y^2}·\sqrt{2}+4y·\sqrt{81}·\sqrt{2}-8·\sqrt{49y^2}·\sqrt{2} \\ 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=3·\sqrt{(8y{)}^2}·\sqrt{2}+4y·\sqrt{81}·\sqrt{2}-8·\sqrt{(7y{)}^2}·\sqrt{2} \\ 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=3·8y·\sqrt{2}+4y·9·\sqrt{2}-8·7y·\sqrt{2} \\ 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=24y\sqrt{2}+36y\sqrt{2}-56y\sqrt{2} \\ 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=(24+36-56)y\sqrt{2} \\ 3\sqrt{128y^2}+4y\sqrt{162}-8\sqrt{98y^2}=4y\sqrt{2} \end{gathered}$$