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On evaluating quadratic equations, It always equals zero:


Why zero? Is it possible to use other number for another purpose?

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Yes, and then you can just subtract the number from both sides..... – user5137 Aug 25 '12 at 21:00
If you use another number, say $d$, instead of $0$, then you can as well study $ax^2 + bx + e = 0$ with $e=c-d$. So this covers every possible case of interest. – user20266 Aug 25 '12 at 21:00
if you want to equate it to another polynomial expression, you can move the right hand side over to the left and you get another expression that gives $P(x)=0$ for a polynomial $P$. This includes cases where you let the right hand side be a constant (which, after moving over, makes another $P(x)=0$ with $P$ being a quadratic expression.) – progressiveforest Aug 25 '12 at 21:02
It's a normal form that we like to study: LHS = 0. In fact, it's trivial to show that we can always take any equation of the form LHS = RHS, and re-write it into that normal form. – user2468 Aug 25 '12 at 21:16
Back in the bad old days when there were no negative numbers outside China and India, Al-Khwarizmi had many types of quadratic equations: $x^2=5x+6$, $x^2+5x=6$, $x^2+6=7x$ were of different types, and required different analyses. But with negative numbers allowed as coefficients, we can make the right-hand side equal to $0$ in all cases, so there is only one type. – André Nicolas Aug 25 '12 at 21:28
up vote 16 down vote accepted

The value of c is a simple number with no variable. So you can move any value on the right side over to the left and it will just become part of c. Example: $$x^2+x-6=6$$ $$x^2+x-12=0$$

Therefore, we can set the right hand side equal to any number we want. We usually set it equal to zero because this helps to solve later. Example: $$(x+3)(x-2) = 6$$ vs $$(x-3)(x+4) = 0$$

The second one is easier to solve because we know anything multiplied by 0 is 0. That means we can solve each part individually.

EDIT: After we reach the factored form, we know the answer is in the form of something multiplied by something else equals a number. If that number is not 0 then we must take both parts into account. On the other hand, if it is 0 then we can simply ask what will make one of those parts zero? Then it doesn't matter what the other part is.
$$(x-3)(x+4) = 0$$ Here, we know that if $(x-3) = 0$ or $(x+4) = 0$ then the whole thing will equal zero, because anything multiplied by 0 is 0. So, we can just ask what value of $x$ will make $(x-3) = 0$ true?

Compare this to the version not set to zero: $$(x+3)(x-2) = 6$$ Now, we can't make this any simpler. We must figure out what value of $x$ will make that entire thing true from the start.

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The second one is easier to solve because we know anything multiplied by 0 is 0. I didn't get it, when will we need to multiply by 0? – Voyska Aug 26 '12 at 11:38
@GustavoBandeira: You have a product of two factors, $(x-3)$ and $(x+4)$, which is $0$. A product of two factors is $0$ iff at least one of the factors is $0$. Thus we know that either $x-3=0$ or $x+4=0$, thus we see immediately that the solutions are $3$ and $-4$. On the other hand, there's no general rule of this sort if the product is $6$ (or any other non-zero number). – celtschk Aug 26 '12 at 20:31

Absolutely! But think about what you end up with. Consider the quadratic equation

$x^2 + 2x + 3 = 2 \, . $

If we now subtract 2 from both sides we get $x^2 + 2x + 1 = 0.$ Meaning that these two equations are just two ways of expressing the same thing. So, to save you the trouble of substracting 2 from both sides, you'll be presented with $x^2 + 2x + 1 = 0$ instead of $x^2 + 2x + 3 = 2.$

In fact, you don't even need a number on the right hand side. What about

$2x^2 + 5x - 9 = x^2 + 3x - 10 \, ? $

I could subtract $x^2 + 3x - 10$ from both sides and end up with our friend $x^2 + 2x + 1= 0$. Any equation of the form $px^2 + qx + r = sx^2 + tx + u$ can be simplified - tidied up, if you will - into the form $ax^2 + bx + c = 0.$ When you come across one in the form $ax^2 + bx + c = 0$ it simply means someone has tidied it all up for you in advance. (And it doesn't change the solutions!)

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DaleSwanson's answer is nice. I just include this as an answer because its too long for a comment:

Consider this, if $a_1,a_2 \neq 0$ then $y=a_1x^2+b_1x+c_1$ and $y=a_2x^2+b_2x+c_2$ give parabolas in the $xy$-plane for particular choices of $b_1,b_2,c_1,c_2$. These parabolas intersect if the equation $a_1x^2+b_1x+c_1 = a_2x^2+b_2x+c_2$ has a solution. Bringing all the terms to the r.h.s yields $(a_2-a_1)x^2+(b_2-b_1)x+c_2-c_1=0$. Let $a=a_2-a_1$, $b=b_2-b_1$ and $c=c_2-c_1$ and we obtain the standard $ax^2+bx+c=0$. Assuming $a \neq 0$ amounts to supposing $a_2 \neq a_1$ and the existence of solutions now characterizes the locations (if any) where the parabolas $y=a_1x^2+b_1x+c_1$ and $y=a_2x^2+b_2x+c_2$ intersect.

More generally, suppose $y=f(x)$ and $y=g(x)$ are graphs of polynomials with $deg(f)=m$ and $deg(g)=n$ and $m<n$ then the number of possible intersections will be at most $n$.

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$$ax^2+bx+c=0 \implies ax^2+bx=-c$$

$$ax^2+bx+c-d=0 \implies ax^2+bx+c=d$$

We generally want the quadratic to equal zero, however, because the solutions are the roots of the quadratic. Roots of functions, i.e. the solutions(s) of functions the form $f(x)=0$ are very important.

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it is general form ,namely second order polynomial equation and express like $f(x)=0$ where $f(x)=a*x^2+b*x+c$ what if this is equal to some number $D$? $a*x^2+b*x+c=D$ so we can write it as $a*x^2+b*x+c-D=0$

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