Please Help SOLVE....?

Question

( ) = radical sign

(p^2 - 3p + 16) = p + 1

Answer 1

Here's how I would do it:

To remove the sqrt, you square both sides:

sqrt(p^2 - 3p + 16) = (p+1)(p+1)
p^2 - 3p + 16 = p^2 + 2p + 1

Then it's basic algebra. The p^2 are out of there, and you are left with

-3p + 16 = 2p + 1
16-1=2p+3p
15=5p
p=3

So, let's check:
sqrt(3^2 - 3(3) + 16) = 3+1
sqrt(9-9+16)=4
sqrt 16 =4

The person who put -5 as one of the answers... it doesn't work:

sqrt ((-5)^2 -3(-5) + 16) = -5+1
sqrt (25+15+16)= -4
sqrt 56= -4 This is false, so -5 is not an answer.

Answer 2

( ) = radical sign

(p^2 - 3p + 16) = p + 1
(p^2 - 3p + 16)- (p+1) = (p +1)- (p+1)
p^2- 3p +16 - p -1 = 0
p^2- 2p + 15 = 0
( p -3 ) ( p+5 ) = 0

p = 3 or p= -5

Answer 3

p^2 - 3p + 16 = p + 1

p^2 - 3p + 16 - p - 1 = 0

p^2 - 3p - p + 16 - 1 = 0

p^2 - 4p + 15 = 0

When everything is changed from right to left, the positive sign (+) changes to negative (-). vice versa.

p^2 - 4p + 15 = 0

hope this helps.

Answer 4

14

Answer 5

sqrt(p^2-3p+16)=p+1
squaring both sides w get,
p^2-3p+16=p^2+2p+1
=>-3p-2p=1-16
=>-5p=-15
=>p=3 ans

Answer 6

use the general law for the square eqation