A few days ago
fairwaydrive322

How do I simplify the square root of -32 minus the square root of -50?

It needs to be in the form a+bi. Thank you!!!!

Top 2 Answers
A few days ago
Anonymous

Favorite Answer

This is a little complicated, so I will explain it the best I can.

I am going to use ‘sqrt’ for the square root symbol.

Simplify:

sqrt(-32) – sqrt(-50) =

sqrt(16 x 2 x -1) – sqrt(25 x 2 x -1)

The sqrt(-1) = i, sqrt(16) = 4, and sqrt(25) = 5, so:

This equals 4i times sqrt(2) minus 5i times sqrt(2) by taking the square root of -1, 16, and 25.

This can also be written:

4isqrt(2) – 5isqrt(2)

Because they both have a base of sqrt(2), you can subract 5isqrt(2) from 4isqrt(2):

5isqrt(2) – 4isqrt(2) = -1isqrt(2)

= -isqrt(2)

To put it in a + bi form:

a – isqrt(2) = a + -isqrt(2)

This means a = 0, so

= 0 + -isqrt(2)

To make it 0+ bi:

The sqrt of 2 times -1 times i

= -sqrt(2)i or 0 + -sqrt(2)i

You can probably leave the 0 out because it doesn’t do anything (unless your teacher needs you to show it), so the answer is:

-sqrt(2)i

I hope this helps. If you need me to explain anything more or if you don’t understand something, let me know and I will try to help. 🙂

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4 years ago
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you will desire to multiply/divide oblong roots sqrt(5)*sqrt(2)=sqrt(10) sqrt(40 2)=sqrt(6)*sqrt(7) so without difficulty reliable out multiples of ‘squares’ then do the arithmetic 6 sq. root 32 – sq. root 50 + oblong root of 8 6(sqrt(sixteen)sqrt(2)) – sqrt(25)(sqrt(2) + sqrt(4)sqrt(2) 6(4sqrt(2)) – 5sqrt(2) + 2sqrt(2) 24sqrt(2) – 5sqrt(2) +2sqrt(2) 21sqrt(2) oblong root of 80 – oblong root of 12 – 5 oblong root of 20 sqrt(4)sqrt(4)sqrt(5) – sqrt(4)sqrt(3) – 5(sqrt4)sqrt(5)) (2)(2)(sqrt(5) – 2sqrt(3) – 5(2sqrt(5)) 4sqrt(5) – 2sqrt(3) – 10sqrt(5) -2sqrt(3)-6sqrt(5)
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