Solving this radical equation

May 2010
6
0
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
Jun 2008
16,217
6,765
North Texas
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?
 
May 2010
6
0
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
Jun 2008
16,217
6,765
North Texas
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?
 
May 2010
6
0
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
Mar 2010
2,340
821
Chicago
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.
 
May 2010
6
0
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
Mar 2010
2,340
821
Chicago
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.