I'm wondering how I solve this kind of equation.

I'm (re)learning 2nd degree equations, and I would have no problem solving $25x^2 - 9$ or $25x^2 - 34x + 9$, for example, but these ones with letters and exponents in the last term are not what I'm used to.

  • $\begingroup$ $$25x^2 - 9y^2 = (5x)^2 - (3y)^2 = (5x+3y)(5x-3y)$$ $\endgroup$ Mar 15, 2016 at 15:26
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    $\begingroup$ Equations feature an equality operator ($=$), I do not see one. $\endgroup$
    – mvw
    Mar 15, 2016 at 15:27
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    $\begingroup$ First of all, $25x^2+0x-9y^2$ is not an equation. You have to "equate" it to something. For example $25x^2-9y^2=0$ is an equation. Then you can solve it with respect to $x$ or respect to $y$. If you solve w.r.t. $x$, you will perform the operations like usual treating $y$ like a number. The rules of math apply in the same way. At the end, the solutions of the equation will contain $y$ as a parameter. This in general. In the case above, assuming your formula is $25x^2-9y^2=(5x+3y)(5x-3y)$ so it will be equal to $0$ when $5x-3y=0$ or when $5x+3y=0$. $\endgroup$ Mar 15, 2016 at 15:31

1 Answer 1


Suppose that the $25x^2 - 9y^2 = 0$ (I took the expression you gave in the title and made an example equation). Here we observe that $25$ and $9$ are perfect squares, so we can rewrite as $(y)^2 = (\frac{5x}{3})^2$. So we see that $y = \pm \dfrac{5x}{3}$. More often than not, there will be a pattern like this. A very useful tool, although simple, is the quadratic formula. In quadratics with two variables, for example $16x^2 + 2xy + 15y^2 = 0$ it is useful to find the value of $x$ or $y$ in terms of the other variable by using the quadratic formula. Another trick is to split up a sum into products and then calling them factors. For example in $m^3 - n^3 = 21$ I can apply the difference of cubes formula and call the subsequent terms factors of $21$ and solve simultaneous equations for each pair of factors.


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