Solving this radical equation

**thecalifornialife** · June 1st 2010, 12:11 PM

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.

**skeeter** · June 1st 2010, 12:16 PM

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?

**thecalifornialife** · June 1st 2010, 12:24 PM

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.

**skeeter** · June 1st 2010, 12:30 PM

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?

**thecalifornialife** · June 1st 2010, 12:30 PM

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?

**undefined** · June 1st 2010, 12:34 PM

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.

**thecalifornialife** · June 1st 2010, 12:37 PM

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

**undefined** · June 1st 2010, 12:43 PM

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.

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