How do you simplify square root of 128 + the square root of 32?

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How do you simplify square root of 128 + the square root of 32?

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Answer 1 · 2018-05-02T18:20:28

$\sqrt{128} + \sqrt{32}$ $sqrt(128)+sqrt(32)$

$\sqrt{(4 \times 4) \times (2 \times 2) \times 2} + \sqrt{(4 \times 4) \times 2}$ $sqrt((4 xx 4) xx (2 xx 2) xx 2) + sqrt((4 xx 4) xx 2)$

$4 \times 2 \sqrt{2} + 4 \sqrt{2}$ $4 xx 2sqrt(2) + 4sqrt(2)$

$8 \sqrt{2} + 4 \sqrt{2}$ $8sqrt(2) + 4sqrt(2)$

$12 \sqrt{2}$ $12sqrt(2)$

Answer 2 · 2018-05-02T18:28:48

$12 \sqrt{2}$ $12sqrt2$

Explanation:

Simplify:

$\sqrt{128} = \sqrt{16} \times \sqrt{8} \to 4 \times \sqrt{4} \times \sqrt{2} = 8 \sqrt{2}$ $sqrt128=sqrt16xxsqrt8 -> 4 xx sqrt4 xx sqrt2=8sqrt2$ .

$\sqrt{32} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$ $sqrt32=sqrt16xxsqrt2=4sqrt2$

$8 \sqrt{2} + 4 \sqrt{2} = 12 \sqrt{2}$ $8sqrt2+4sqrt2=12sqrt2$

Answer 3 · 2018-05-02T18:41:45

$\sqrt{128} + \sqrt{32} = 2 \cdot \sqrt{12}$ $\sqrt(128)+\sqrt(32)=2*\sqrt(12)$

Explanation:

$\sqrt{128} + \sqrt{32}$ $\sqrt(128)+\sqrt(32)$
Write both numbers as a product of prime numbers.
(unique factorisation ).

For example:
my pic

$128 = 2 \cdot 64 = 2 \cdot 2 \cdot 32$ $128=2*64=2*2*32$
$32 = 2 \cdot 16 = 2 \cdot 2 \cdot 8 = 2 \cdot 2 \cdot 2 \cdot 4 = 2^{5} = 2^{4} \cdot 2$ $32=2*16=2*2*8=2*2*2*4=2^5=2^4*2$

$\sqrt{128} + \sqrt{32} = \sqrt{2 \cdot 2 \cdot 32} + \sqrt{32} =$ $\sqrt(128)+\sqrt(32)=\sqrt(2*2*32)+\sqrt(32)=$
$= \sqrt{2 \cdot 2} \cdot \sqrt{32} + \sqrt{32} = 2 \cdot \sqrt{32} + \sqrt{32} =$ $=\sqrt(2*2)* \sqrt(32)+\sqrt(32)=2* \sqrt(32)+\sqrt(32)=$
$= 3 \cdot \sqrt{32} = 3 \cdot \sqrt{2^{4} \cdot 2} = 3 \cdot \sqrt{2^{4}} \cdot \sqrt{2} = 3 \cdot 2^{2} \cdot \sqrt{2} =$ $=3*\sqrt(32)=3* \sqrt(2^4*2)= 3*\sqrt(2^4)*\sqrt(2)= 3*2^2*\sqrt(2)=$
$= 3 \cdot 4 \cdot \sqrt{2} = 12 \sqrt{2}$ $=3*4*\sqrt(2)=12\sqrt(2)$

Answer 4 · 2018-05-02T19:16:56

$12 \sqrt{2}$ $12sqrt(2)$

Explanation:

Given: $\sqrt{128} + \sqrt{32}$ $sqrt(128)+sqrt(32)$

You are looking for squared values that you can 'extract' from the roots.

If you are ever not sure do a quick sketch of a prime factor tree to help:
Tony B

$\sqrt{128} + \sqrt{32}$ $sqrt(128)+sqrt(32)$

$\sqrt{2^{2} \times 2^{2} \times 2^{2} \times 2} + \sqrt{2^{2} \times 2^{2} \times 2}$ $sqrt(2^2xx2^2xx2^2xx2)+sqrt(2^2xx2^2xx2)$

$ddddd 8 \sqrt{2} dddddd + dddd 4 \sqrt{2} dddd = d 12 \sqrt{2}$ $color(white)("ddddd")8sqrt(2)color(white)("dddddd")+color(white)("dddd")4sqrt(2)color(white)("dddd")=color(white)("d")12sqrt(2)$

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How do you simplify square root of 128 + the square root of 32?

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