Hi, This is perhaps the most rookie question to Mathematics that one could ask, but I have searched for information on the question and found only one source which contained an answer. The answer was that it is because otherwise we wouldn't know what the value of either of the factors is. I must therefore be confused. We know that 1×a=a, thus (x+1)(x+1)=1 if both of the factors equal 1. In this case, if we set x=0 then we have (0+1)(0+1)=1; 1×1=1, therefore our factor is 1. As far as I know, there is no other value of x which can yield a value of 1, so why do we set the equations equal to 0? Thanks for any help.
First, quadratic equations are NOT necessarily set equal to 0. That is one way of solving a quadratic equation because then if we can factor we can use the "zero product property": if ab= 0 then either a= 0 or b= 0. If ab equals any number other than 0, that there are many ways to factor ab. That is true because 0 has the special property that any number times 0 is equal to 0. Since that is true, a is not 0, we can multiply both sides of ab= 0 by 1/a to get (1/a)(ab)= (1/a)(0): b= 0. Then you haven't tried very hard x= -1 also works. But that is a square and, as I said, dealing with squares you don't have to set it equal to 0. But for non-squares, (x+ 1)(x+ 2), say, you cannot do that. If (x+1)(x+2)= 1, we cannot say that x+ 1= 1 and x+ 2= 1. But if I were going to solve a quadratic equation by 'completing the square', I would want it written [itex]ax^2+ bx= c[/itex], not [itex]ax^2+ bx+ c= 0[/itex].
You have a point there, even though you didn't explore all the solutions. If the article said " because otherwise we wouldn't know what the value of either of the factors is", that's too sweeping a statement. I suspect the article was trying to convey that an equation like (A)(B) = 4 doesn't narrow down the possibilities for A and B down to a finite number of choices. It meant that knowing the product of two factors is a non-zero number and not knowing anything else about the factors (such as the fact they are equal to each other) doesn't let you narrow down the possibilities for their values to a finite set of numbers.
If x=-1 then (-1+1)(-1+1) would equal 0 rather than 1, but I see your point: If I had (x+2)(x+3)=1 it would be impossible to solve since there is no number (x+2)=1/(x+3) or (x+3)=1/(x+2) but there is a number where (x+2)=0 and (x=3)=0. Would you mind showing me the mathematical proof that the former statement is true so that I may have a sound mind?
The simple answer to your question is that so you can find the roots. It is very common to need to know when an equation (quadratic or other) is equal to zero. That is why you set it to zero and solve. BTW, try x=-1.382 or x = -3.62 in your equation.
I found that those values came close to 1, I suppose there is a number of infinite decimal places that satisfies the equation then. Ok, so it is possible to find roots that satisfy the above equation. When you calculated the above did you first convert the factors into their second degree form by calculating their product and subtracting 1? Why is it common to need to know when the equation is equal to 0? What are it's uses? I am beginning to enter confusion once again. I really wish to understand where the idea of setting equations equal to 0 comes from and thus why we do it. At the moment it is just a mechanical process in my mind - it is what we do - there is not history to it. Algebra typically presents us with an expression like x^{2}+bx+c and we wish to find what x is so we set the expression equal to 0. My confusion is why 0 is used and not 1 for example; as you showed above, there is a number that satisfies it, although an irrational number.
Yes, I wrote the equation in standard form (ax^{2}+bx +c) then subtracted 1 from each side and found the roots of the new equation. There are a variety of reasons why you need to find the roots of an equation, note that this is true whether or not the equation is quadratic, or even polynomial. The roots of a derivative of a equation give you the points where the original equation is horizontal. For instance your equation x^{2}+5x+6 has a derivative of 2x + 5. Finding the root of 2x+5=0 => x =-5/2 gives you the horizontal point of the quadratic, which is the minimum value. and yes, I rounded off the the values for x^{2}+5x+6=1 try -1.381966011 and -3.618033989
Yes, that would be the easiest approach, or even having used a graphing calculator or whatnot. Because if we have two numbers multiplied together to equal 0, so AB=0, then either the first is equal to 0, or the second (or both). But when two numbers are multiplied together to equal something else, say AB=1, then we aren't getting any information from A and B because they relate to each other in some way. If we chose A=1, then B must be equal to 1 (we would have to see if it satisfies our equation). But we could also choose A=1/2 and then B must be 2. If this combination doesn't work either, try again. A=-50, B=-1/50 etc. We would just be testing all possible A and B anyway, which is no different to guessing x to find the solution to a quadratic. Now, of course it's a different story when we have the equation set equal to 0. No guessing required, we just know that either of the factors (or both) must be equal to 0.
That may be your impression, but it's not what usually happens. Algebra, in the sense of applications of algebra, usually presents us with equations where the right hand side may or may not be zero. It is possible to convert such an equation to an equation whose right hand side is zero. This is often done. The reason it is often done is because of what we've been discussing. If you can factor an expression that is equal to zero you can try to find soluitons by setting each factor equal to zero. When you say "Algebra typically presents us with an expression...", I think you mean that you are often presented with a function like [itex] f(x) = a x^2 + bx + c [/itex]. There are various practical reasons why people are interested in the values of [itex]x[/itex] where a function is zero. You could also ask for a value of [itex] x [/itex] where [itex] f(x) [/itex] was equal to 15.