Thread: Solving this radical equation

How do I solve this radical equation? (the sqrt stands for square root)
$\displaystyle sqrt(x-3)+1=x$


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

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

Originally Posted by thecalifornialife

How do I solve this radical equation? (the sqrt stands for square root)
$\displaystyle sqrt(x-3)+1=x$


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

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

$\displaystyle b^2-4ac < 0$ ... what's that tell you?

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.

Originally Posted by thecalifornialife

How do I solve this radical equation? (the sqrt stands for square root)
$\displaystyle sqrt(x-3)+1=x$


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

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?

However, if I plug the equation to the quadratic formula, I get 1$\displaystyle /2(3+sqrt(7i))$

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?

Originally Posted by thecalifornialife

However, if I plug the equation to the quadratic formula, I get 1$\displaystyle /2(3+sqrt(7i))$

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.

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

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

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

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

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

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

The $\displaystyle i$ should indeed be outside the radical.