# First degree or second degree quadratic equation?

$X-\sqrt{121}= 0$

$Y^2-121=0$

When I try to solve by squaring on both sides the first equation changes to $X^2-121 =0$, where $x = \pm 11$.

When I try to solve it this way:

$x= \sqrt{121} = +11$, there is only one solution.

Why the difference?

• Please use MathJax. Aug 29, 2014 at 11:30
• actually it's unreadable !! For the first equation, is it $x-\sqrt{121}=0$ ? Then, why squaring on both side ? you immediately have $x=11$. Indeed, $$x-\sqrt{11}=0\implies x-11=0\implies x=11.$$ For the other one, is it $y^2-121=0$ ? If yes, then $y^2=121\implies y=\pm\sqrt{121}=\pm 11$
– idm
Aug 29, 2014 at 11:37

The equation $x=a$ has just one solution.
$x^2=a^2$ is equivalent to $0=x^2-a^2=(x+a)(x-a)$ with solutions $x=a$ and $x=-a$.
Likewise taking the square root (using the convention that the positive real root is taken) can lose solutions of the original equation. This is why the general solution to the quadratic is written with $\pm$ before the square root.