Respuesta :
If a quadratic equation has solutions, the standard procedure will always work: given an equation like
[tex] ax^2+bx+c=0 [/tex]
The solutions (if any) are given by
[tex] x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
In your case, you have
[tex] a=-9,\ b=0,\ c=1 [/tex]
So the formula becomes
[tex] x_{1,2} = \dfrac{\pm\sqrt{36}}{18}= \pm\dfrac{6}{18}=\pm\dfrac{1}{3} [/tex]
Anyway, this is quite a special case, because you're missing the linear term (since [tex] b=0[/tex])
This means that you can solve equations like these more easily: rearrange the equation to make it look like [tex] ax^2=c [/tex]. In your case, it becomes
[tex] 9x^2=1[/tex]
Divide both sides by 9:
[tex] x^2=\dfrac{1}{9} [/tex]
Now, you know that the square of a number equals 1/9. By definition, it means that this number is the square root of 1/9. Nevertheless, both the square root and its opposite are solutions of the equation, because the minus sign will cancel out when squaring.
So, in general, you have
[tex] ax^2=c \iff x^2 = \dfrac{c}{a} \iff x=\pm\sqrt{\dfrac{c}{a}[/tex]
which of course makes sense, if you're using real numbers, only if c/a>0.
In your case, this becomes
[tex] 9x^2=1 \iff x^2 = \dfrac{1}{9} \iff x=\pm\sqrt{\dfrac{1}{9}} = \pm\dfrac{1}{3} [/tex]
