[Math] what does “solve the equation for x” mean

algebra-precalculus

I'm not expert for the mathematics, and familiar with this.
But when I learn the math in high school, I've got the sort of the below equation things from teacher or books.

Solve the equation for $x$:
$$(x^2 + 6x -7)(2x^2 – 5x-3) = 0$$

After $10$ years, now I'm looking the math book again.
Especially, I have one question what I don't know.

  1. Why? Do we have to solve that equation which has equal to 'zero' and what does it mean some equation is the same to zero?
    why do we set the zero to equation to solve that sort of equations?

Best Answer

"Solve the equation .... for $x$" simply means that you figure out for which value(s) of $x$ the equation holds true. So, for example, if I say:

"Solve the equation $2x=6$ for $x$", then you should say: "Ah, yes, that equation holds true for $x=3$". Why?

Because if you fill in $x=3$ you indeed get $2x = 2\cdot3 = 6$.

The equation would not hold for $x=4$, since for $x=4$ we get $2x = 2\cdot4 = 8 \not = 6$

(indeed, you can show that $x=3$ is the only value for $x$ that would make $2x=6$ true)

Also, notice that an equation can hold for multiple values of $x$. For example, take $x^2 = 1$. That equation holds true for $x = 1$, but also for $x = -1$. But it holds false for any other value of $x$.

Finally, notice that there need not be any $0$'s involved with these kinds of problems: there were no $0$'s in the two examples I just gave.

However, we often like to set things equal to $0$, because if you have something like what you have:

$$(x^2 + 6x -7)(2x^2 - 5x-3) = 0$$

then we can 'divide and conquer', since this equation will hold true if either

$$(x^2 + 6x -7) = 0$$ or

$$(2x^2 - 5x-3) = 0$$

and so now I just need to 'solve' those two simpler equations. In other words, setting one side of the equation to $0$ will often simplify matters. So, for example, I could have taken $2x=6$ and changed that into $2x-6=0$. And $x^2=1$ can be changed into $x^2-1=0$

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