Question

Bonjour pourriez vous m'aider avec cette équation du second degré? j'ai du mal à comprendre le concept. merci

Answer 1

If a quadratic equation has solutions, the standard procedure will always work: given an equation like

$ax^2+bx+c=0$

The solutions (if any) are given by

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your case, you have

$a=-9,\ b=0,\ c=1$

So the formula becomes

$x_{1,2} = \dfrac{\pm\sqrt{36}}{18}= \pm\dfrac{6}{18}=\pm\dfrac{1}{3}$

Anyway, this is quite a special case, because you're missing the linear term (since $b=0$)

This means that you can solve equations like these more easily: rearrange the equation to make it look like $ax^2=c$. In your case, it becomes

$9x^2=1$

Divide both sides by 9:

$x^2=\dfrac{1}{9}$

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

$ax^2=c \iff x^2 = \dfrac{c}{a} \iff x=\pm\sqrt{\dfrac{c}{a}$

which of course makes sense, if you're using real numbers, only if c/a>0.

In your case, this becomes

$9x^2=1 \iff x^2 = \dfrac{1}{9} \iff x=\pm\sqrt{\dfrac{1}{9}} = \pm\dfrac{1}{3}$