Question

What is the extraneous solution to sqrt 2p+1+2 sqrt p=1

Answer 1

p = 4

Explanation:

Answer 2

Given equation:
`√(2p+1)+2√(p)=1`.

`\mathrm{Subtract\:}2√(p)\mathrm{\:from\:both\:sides}`

`√(2p+1)+2√(p)-2√(p)=1-2√(p)`

`√(2p+1)=1-2√(p)`

`\mathrm{Square\:both\:sides}`

`\left(√(2p+1)\right)^2=\left(1-2√(p)\right)^2`

`2p+1=1-4√(p)+4p`

`\mathrm{Subtract\:}1+4p\mathrm{\:from\:both\:sides}`

`2p+1-\left(1+4p\right)=1-4√(p)+4p-\left(1+4p\right)`

`-2p=-4√(p)`

`\left(-2p\right)^2=\left(-4√(p)\right)^2`

`4p^2=16p`

`4p^2-16p=0`

Factoring out 4p, we get

`4p(p-4) =0`

4p=0 and p-4 =0

p=0 and p=4.

`\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}√(2p+1)+2√(p)=1`

`\mathrm{Plug}\quad p=4:\quad √(2\cdot \:4+1)+2√(4)=1\quad \Rightarrow \quad \mathrm{False}`

`\mathrm{Plug}\quad p=0:\quad √(2\cdot \:0+1)+2√(0)=1\quad \Rightarrow \quad \mathrm{True}`

Therefore, x=4 is the extraneous solution to the given equation.