How do you evaluate sqrt32 + 2sqrt18 - 3sqrt50 + 3sqrt98?

3 Answers
Mar 16, 2018

16sqrt2

Explanation:

sqrt16 timessqrt 2 = sqrt32
2 times sqrt9 times sqrt2 = 2sqrt18
-3 times sqrt25 times sqrt2 = -3sqrt50
3 times sqrt49 times sqrt2 = 3sqrt98

4sqrt2 = sqrt32
6sqrt2 = 2sqrt18
-15sqrt2 = -3sqrt50
21sqrt2 = 3sqrt98

4+6-15+21 = 16sqrt2
You can combine them because they all have sqrt2

Mar 16, 2018

See below

Explanation:

sqrt32+2sqrt18-3sqrt50+3sqrt98=sqrt(8·4)+2sqrt(9·2)-3sqrt(10·5)+3sqrt(49·2)=sqrt(2^2·2·2^2)+2sqrt(3^2·2)-3sqrt(5^2·2)+3sqrt(7^2·2)=

Due to this obvious identity we can extract some factor from square root simbol (this indentity is sqrt(a^2)=a)

4sqrt2+6sqrt2-15sqrt2+21sqrt2=16sqrt2

Mar 16, 2018

=16sqrt2

Explanation:

First write each radicand as the product of its factors, using square numbers wherever possible,

sqrt32" " +" " 2sqrt18" " -" " 3sqrt50" " +" " 3sqrt98

=sqrt(16xx2)" " + 2sqrt(9xx2)" " - 3sqrt(25xx2)" " + 3sqrt(49xx2)

Find the square roots where possible

=4sqrt2" " +" " 2xx3sqrt2" "-" " 3xx5sqrt2" + " 3xx7sqrt2

=4sqrt2" " +" " 6sqrt2" "-" " 15sqrt2" + " 21sqrt2

Now you can simplify because they all have sqrt2

=16sqrt2