#### 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.

#### skeeter

MHF Helper
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?

#### 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.

#### skeeter

MHF Helper
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?

#### 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?

#### undefined

MHF Hall of Honor
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.

#### thecalifornialife

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

#### undefined

MHF Hall of Honor
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.