Question

simplify 4 root 2 + 3 root 32

Answer 1

$\huge\green{Heya.....}$

$4 \sqrt{2} + 3 \sqrt{32} \\ 4 \sqrt{2} + 3 \sqrt{16 \times 2} \\ 4 \sqrt{2 } + 3 \times 4 \sqrt{2} \\ 4 \sqrt{2} + 12 \sqrt{2} \\ 4 + 12 \sqrt{2} \\ 16 \sqrt{2}$

Hope it helps uhhh....

Answer 2

Answer:

After simplification value of the given term is $16 \sqrt{2} .$

Step-by-step explanation:

Given,4 root 2 + 3 root 32

i.e $4 \sqrt{2} + 3 \sqrt{32}$

Here first we need to break 32.

$32 = {2}^{5} = 2 \times 2 \times 2 \times 2 \times 2$

Now $4 \sqrt{2} + 3 \sqrt{32} = 4 \sqrt{2} + 3 \sqrt{2 \times 2 \times 2 \times 2 \times 2}$

$= 4 \sqrt{2} + 3 \times 2 \times 2 \sqrt{2}$

$= 4 \sqrt{2} + 12 \sqrt{2}$

$= 16 \sqrt{2}$

After simplification value of the given term is $16 \sqrt{2} .$

This is a problem of power of indices.

Here applied formulas are$x \sqrt{y} + z \sqrt{y} = (x + z) \sqrt{y}$

$\sqrt{x \times x} = x$

$\sqrt{x \times x \times y} = x \sqrt{y}$

Some others formula of power of indices are

${x}^{y} \times {x}^{z} = {x}^{z + y}$

${x}^{1} = x$

${x}^{0} = 1$

${y}^{x} \times {z}^{x} = {(yz)}^{x}$

e.t.c.