• Jun 1st 2010, 01:11 PM
thecalifornialife
How do I solve this radical equation? (the sqrt stands for square root)
$sqrt(x-3)+1=x$

$sqrt(x-3)=x-1$
$(sqrt(x-3))^2=(x-1)^2$
$x-3=(x-1)(x-1)$
$x-3=x^2-x-x-1$
$x-3=x^2-2x+1$
$0=x^2-3x+4$

I need an actual number so I can check the solution.
• Jun 1st 2010, 01:16 PM
skeeter
Quote:

Originally Posted by thecalifornialife
How do I solve this radical equation? (the sqrt stands for square root)
$sqrt(x-3)+1=x$

$sqrt(x-3)=x-1$
$(sqrt(x-3))^2=(x-1)^2$
$x-3=(x-1)(x-1)$
$x-3=x^2-x-x-1$
$x-3=x^2-2x+1$
$0=x^2-3x+4$

I need an actual number so I can check the solution.

$b^2-4ac < 0$ ... what's that tell you?
• Jun 1st 2010, 01:24 PM
thecalifornialife
Um.... i'm not sure how to connect the two equations. If I use the equation with what is left over, I get 7>0.
• Jun 1st 2010, 01:30 PM
skeeter
Quote:

Originally Posted by thecalifornialife
How do I solve this radical equation? (the sqrt stands for square root)
$sqrt(x-3)+1=x$

$sqrt(x-3)=x-1$
$(sqrt(x-3))^2=(x-1)^2$
$x-3=(x-1)(x-1)$
$x-3=x^2-x-x-1$
$x-3=x^2-2x+1$
$\textcolor{red}{0=x^2-3x+4}$

you did everything correctly, and the last equation has no real solution.

what does that tell you about any real solutions for the original equation?
• Jun 1st 2010, 01:30 PM
thecalifornialife
However, if I plug the equation to the quadratic formula, I get 1 $/2(3+sqrt(7i))$

The solver on quickmath.com says the I is outside the square root. I dont know how they got that. But.... is that the solution?
• Jun 1st 2010, 01:34 PM
undefined
Quote:

Originally Posted by thecalifornialife
However, if I plug the equation to the quadratic formula, I get 1 $/2(3+sqrt(7i))$

The solver on quickmath.com says the I is outside the square root. I dont know how they got that. But.... is that the solution?

The equation has no real solutions, only imaginary (complex) ones. If you're looking for complex solutions, please specify.
• Jun 1st 2010, 01:37 PM
thecalifornialife
Is the imaginary solution I get incorrect? I would hate to teach my girlfriend this for her test and she gets it wrong.
• Jun 1st 2010, 01:43 PM
undefined
Quote:

Originally Posted by thecalifornialife
Is the imaginary solution I get incorrect? I would hate to teach my girlfriend this for her test and she gets it wrong.

Okay, so

$x^2-3x+4=0$

$x=\frac{3\pm\sqrt{9-16}}{2}$

$x = \frac{3\pm\sqrt{-7}}{2}$

$x = \frac{3\pm i\sqrt{7}}{2}$

The $i$ should indeed be outside the radical.