Question

Subtract the following radicals: 5√45 - 3√5
A. 5√5
B. 12
C. 12√5
D. 42√5

Answer 1

@dan815

Answer 2

@Preetha

Answer 3

@SolomonZelman

Answer 4

\(\large {
5\sqrt{45}-3\sqrt{5}\qquad {\color{brown}{ 45\to 3\cdot 3\cdot 5\to 3^2\cdot 5 }}
\\ \quad \\
5\sqrt{45}-3\sqrt{5}\implies 5\sqrt{{\color{brown}{ 3^2\cdot 5}}}-3\sqrt{5}\implies ?
}\)

Answer 5

How would I find that answer?

Answer 6

take out what you can from the 1st radical
what do you have left?

Answer 7

9*5 is 45, so its 5 sqrt 45?

Answer 8

well... the 3 comes out WITHOUT the exponent
thus \(\large {
5\sqrt{45}-3\sqrt{5}\qquad {\color{brown}{ 45\to 3\cdot 3\cdot 5\to 3^2\cdot 5 }}
\\ \quad \\
5\sqrt{45}-3\sqrt{5}\implies 5\sqrt{3^2\cdot 5}-3\sqrt{5}\implies 5\cdot 3\sqrt{5}-3\sqrt{5}
\\ \quad \\
15{\color{blue}{ \sqrt{5} }}-3{\color{blue}{ \sqrt{5} }}\implies ?
}\)

Answer 9

Oh okay uhm its prime isnt it?

Answer 10

the 5 is prime.... the 3 was squared
notice the root is 2, so the 3 is raised at the "2"
the root is at the "2"
so
the 3 comes out without its exponent
but yes, the 5 is prime, stays inside

Answer 11

Ok what next

Answer 12

so if we made say \(\bf \sqrt{5} = a\)
15a - 3a = ?

Answer 13

12

Answer 14

yeap

thus \(\large 15{\color{blue}{ \sqrt{5} }}-3{\color{blue}{ \sqrt{5} }}\implies 12{\color{blue}{ \sqrt{5}}}\)

Answer 15

Thanks!