Is this right?

[debbie29]

Can anyone tell me if I'm doing this right?

sqrt 2x-1 - sqrt x+3 =1

sqrt 2x-1 =1- sqrt x+3

(sqrt 2x-1)^2 = (1- sqrt x+3)^2

2x-1= 1+2 sqrt x+3+ (x-3)

x-3= 2 sqrt x+3

(x-3)^2 = (2 sqrt x+3)^2

x^2- 6x+9= 4(x-3)

x^2-6x+9= 4x-12

x^2- 10x +21=0

(x-7) (x-3) = 0

x-7+0 x-3=0
x=7 x=3

solution: x=7 or x=3

 

[soroban]

Hello, debbie29!

Did you check your answers? . Neither of them work!

And parentheses would help us read your stuff . . .

sqrt(2x-1) - sqrt(x+3) =1

sqrt(2x-1) = 1 - sqrt(x+3)

[sqrt(2x-1)]^2 = [1- sqrt(x+3)]^2

2x - 1 = 1 + 2sqrt(x+3)+ x - 3 . <--- error!

It should be: .2x - 1 . = . 1 + 2sqrt(x+3) + x + 3

And we have: .x - 5 . = . -2sqrt(x+3)

Square: .[x - 5]² . = . [-2sqrt(x+3)]²

. . . x² - 10x + 25 . = . 4(x + 3)

. . . x² - 14x + 13 . = . 4x + 12

. . . (x - 1)(x - 13) . = . 0

Hence: .x = 1, 13

But x = 1 is extraneous . . . it doesn't check.

Therefore, the only solution is: x = 13

 

[debbie29]

Thanks!! I see how to do that now!!!

sorry-- not too good at typing out math. I'll remember to use parentheses next time!!

THANK YOU!

Deb