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?
  • $\begingroup$ When you can factor some expression, and the two factors multiply together to make zero, then that means that one of the factors must equal zero. $\endgroup$ – Doug M Apr 20 '17 at 16:16
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    $\begingroup$ "Solve the equation for $x:~~~~~~(x^2+6x-7)(2x^2-5x-3)=0$" can be reworded to "Find the set of values of $x$ for which the equation $(x^2+6x-7)(2x^2-5x-3)=0$ is true" If you plug in $1$ for example, you would find it simplifies to $(1^2+6\cdot 1-7)(2\cdot1^2-5\cdot1-3)$ which simplifies to $(0)\cdot(-6)$ which is indeed equal to zero so $1$ is indeed in the "set of solutions", but if you were to plug in something like $1000$ you will end up with a large nonzero number somewhere around $2\cdot 10^{12}$ so $1000$ is not in our set of solutions. $\endgroup$ – JMoravitz Apr 20 '17 at 16:16
  • $\begingroup$ @DougM Thanks But I've got still question. why that factor should be zero? $\endgroup$ – excelman Apr 20 '17 at 16:18
  • $\begingroup$ What is zero times anything? $\endgroup$ – Ian Miller Apr 20 '17 at 16:19
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    $\begingroup$ What exactly is troubling you? $(x^2 + 6x - 7)(2x^2 -5x -3)$ is a rule that if you given some number as input you get some number as output. If your inpupt is 2, your output is -45. ($x=2; (2^2+6*2-7)(2*2^2 - 5*2 -3) = -45$). If you input is 3, your output is $(3^2 + 6*3 -7)(2*3^2 - 5*3 -3)= 20*0 = 0$. etc. So the question "when is $(x^2 + 6x -7))(2x^2 -5x -3) = 0$" is "what numbers can $x$ be that will result in $(x^2 + 6x -7))(2x^2 -5x -3) =0$". And.... I don't understand where your confusion lies. $\endgroup$ – fleablood Apr 20 '17 at 16:38

If the equation is just in terms of $x$ and no other variable (like yours), it usually means to find every value of $x$ that, when put into that equation, is true. So, for yours, that means find all values of $x$ such that when I put it into the left hand side $$(x^2 + 6x -7)(2x^2 - 5x-3)$$ I get the right hand side ($0$ in this case). If there are other variables, like $y=2x+4$. it means to change it into the form $$x=\text{anything except $x$}$$ (although this cannot always be done). This is useful for plotting things because say we had $x=y^2+y+1$ and we want to test some points to see what it looks like, we can just plug in a $y$ value and immediately get an $x$ value.

Now, for your equation, for it to be true, we need either $x^2+6x-7=0$ or $2x^2-5x-3=0$ (because anything multiplied by zero would get you $0$ on the right hand side. So, find all $x$ that do either of those two things. For $x^2+6x-7$, we can factor it into $(x+7)(x-1)=0$. So, it looks like $x=-7$ or $x=1$ would make that zero and hence make our original equation satisfied. The same goes for $2x^2-5x-3=(2x+1)(x-3)\implies 2x+1=0$ or $x-3=0$. The first equation we can "solve for $x$" to see that $x=-\frac{1}{2}$ works and for the second, we see that $x=3$ works.

Therefore, any one of $x=-\frac{1}{2},x=3,x=-7$, or $x=1$ for satisfy this equation and we have "solved for $x$".


I will take it with a simpler example for the ease of illustration and concentration on the main issue. So what does it mean by solving $x+3=5$ for $x$? It means to find something that plus $3$ equals $5$. Since equations can be complicated, mathematicians developed symbols to less the burden. Consider instead the problem of finding two numbers of the kind that the first one plus $2$ equals $4$ and that the second one plus $1$ equals $8$. Wow, it is so cumbersome in that form right? Let us see how symbols simplify the words: find some $x,y$ such that $x+2 = 4$ and $y + 1 = 8$; or, in a more usual language: solve the equations $x+2=4$ and $y+1=8$ for $x,y$. You have seen that symbolic language is not supposed to be a confusing thing at all as long as one can correctly translate it!

So solving an equation is more or less like solving a puzzle. For example, consider the puzzle: find a person such that he or she requires others to have higher moral standard than himself or herself (by the way, every human being is a "solution" to this puzzle). If you want, you can rephrase this puzzle as something like: find some person $x$ such that $x$ requires others to have higher moral standard than $x$, or like: solve "$x$ requires others to have higher moral standard than $x$" for $x$. (Rigorously speaking, here we want to specify further the range of $x$; we want to solve the requirement pertaining to $x$ for $x$ being a person.)


"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 $x^2-1=0$

  • $\begingroup$ Thanks can you explain me some case what " this equation will hold false"? actually, I can't get it the meaning that. how to know that equation whether will hold true or not? $\endgroup$ – excelman Apr 20 '17 at 16:32
  • $\begingroup$ For the smaller example of $2x=6$, if you plug in $3$ in place of $x$, you get $2\cdot 3 = 6$ which is a true statement, the left side simplifies to $6$ and the right side is $6$ itself and $6$ is certainly equal to $6$. However, if you were to plug in a different value for $x$, say for example $5$, you would get $2\cdot 5=6$ which is not true. The left simplifies to $10$ while the right is equal to $6$, but $10$ and $6$ are different numbers. We have $10\color{red}{\neq}6$, so $10=6$ is considered false (i.e. not true). $\endgroup$ – JMoravitz Apr 20 '17 at 16:35
  • $\begingroup$ @excelman Updated my answer ... $\endgroup$ – Bram28 Apr 20 '17 at 16:44
  • $\begingroup$ "this equation will hold false" means... the equation is not true. "1 + 1 = 3" is false. $(7^2 + 6(7) - 7)(2*7^2 - 5*7 - 3) = 0$ is not true. So if $x=7$ then $(x^2 + 6(x) - 7)(2*x^2 - 5*x - 3)=(7^2 + 6(7) - 7)(2*7^2 - 5*7 - 3)= 0$ is false. So if we are told "$x$ is some number. $(x^2 + 6(x) - 7)(2*x^2 - 5*x - 3)=0$; what number(s) can $x$ be?" it is impossible that $x$ is $7$ because $(7^2 + 6(7) - 7)(2*7^2 - 5*7 - 3)= 0$ is simply.... not true. $\endgroup$ – fleablood Apr 20 '17 at 16:46

There are some good answers here so far, but none of them really answer the "Why?" part of the question. Why do mathematicians take functions (especially quadratic functions), set them to zero, and solve for $x$? Because the values of $x$ that make the function $0$ are called the roots (or zeroes) of the function. The roots are the values that correspond exactly with where the function crosses the $x$-axis.

Why is that important? Without having to graph the function (which is easier today than it was before the internet and handheld graphing calculators), you can determine some of the function's behavior by finding its roots. A skilled mathematician could find the roots to your sample equation in their head, and so with little effort they can find out a lot about the function.

A real-world example. Let's say you are a physicist calculating where your projectile will land. You have a quadratic function that models the projectile's flight. When does the projectile land? When the function is zero. So you set the function equal to $0$ and solve for $x$.


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