# Equations and Solving Them

I am trying to answer some problems regarding simple equations and rational numbers, and I have an algebra book, but it shows me nothing. It does explain however that common factors are not equal to zero... Though that doesn't teach me much. Could someone please help me?

The equation is this:

$$49x^2 - 98x + 40 = 0$$

Thank you.

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Welcome to MSE =D –  Patrick Da Silva Nov 20 '11 at 0:59
In your comment to Patrick's answer you mentioned that you've found a factorization of the left-hand side. Why didn't you mention that in your question? –  Henning Makholm Nov 20 '11 at 1:17
If you have found a factorization of the left side, then it is not clear what your question is. What do you need help with? –  Gerry Myerson Nov 20 '11 at 11:49

There is a canonical formula for solving a quadratic equation, and it is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ when $p(x) = ax^2 + bx + c$. The reason for this is that $$a\left( x - \left( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right) \right) \left( x - \left( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right)\right) = ax^2 + bx + c.$$ (I leave the fact that this is true for you as an exercise. Just expand the product.) Thus, the solutions to your equation are $$\frac{98 \pm \sqrt{98^2 - 4 \cdot 40 \cdot 49}}{2 \cdot 49}.$$ This formula is very well known and very useful so you should remember it if you are going to do mathematics for a while.
@mth143 : An equation that is not written in the form $ax^2 + bx + c=0$ can also be said to be a quadratic equation if we can manage to write it in that form. Your example is also a quadratic equation. If you used the above formula to factor it, you would've got $49(x-4/7)(x-10/7) = 0$. –  Patrick Da Silva Nov 20 '11 at 1:22