Why the roots of any equation are searched at $0$ point?

Question

I'm trying to understand why we are searching roots of any equation at 0.

For example, $x^2 + 8x - 425 = 0$

The given equation has $2$ roots: $x_1 = -25$ and $x_2 = 17$. It means that the plot of the given function intersects $Ox$ at these points.

But why can't we search for $x$ arguments at 1 point for example. I mean, $x^2 + 8x - 425 = 1$. The roots will be different for the given function result with these arguments, but the function remains the same.

What I want to say, we can always find arguments for the given function result (that's why we can draw a plot of this function). There is always a match between arguments and function result. We can always make a correct numerical equality with different arguments (though the arguments won't be always integers).

Why roots are considered as roots only at the point where a plot intersects $Ox$?

Apologize if this is a simple question, seems like I missed that concept.

Answer 1

Well, if you want to find the intersection points between a parabola, $y= ax^2+bx+c$, with a line $y=dx+e$ (special case the horizontal line $y=d$, say $y=1$ as above), or another parabola, $y= fx^2+gx+h$, you just set the two equations equal:

$ax^2+bx+c = dx+e$ and $ax^2+bx+c=fx^2+gx+h$

and then set them zero:

$ax^2+(b-d)x+(c-e)=0$ and $(a-f)x^2+(b-g)x+(c-h)=0$.

Then you have to determine the roots of these equations.

Answer 2

First we use 0 because we want points(roots) that are on x-axis ($y=0$) and equation ($ax^2 + bx + c =y$) .We substitute y in the equation.

Second we have formula and complete square method to determine roots of any quadratic equation that works only on standard form $(ax^2 +bx+c)$.So we bring equation into its standard form then solve for $x$

If we were to find intersection points of two graphs $2x^2+4x-5 =y$ and $-4x^2+2x+3 =y $ Then what we would do is substitute y from one equation to another equation

So this becomes $-4x^2+2x+3 =2x^2+4x-5 $

We know quadratic formula but that works only on standard form se we bring this equation to its standard form

$-6x^2 -2x +8 =0$

Now we can use quadratic formula to solve for $x$.This gets us points of intersection.

Answer 3

Assume you have a polynomial equation of degree n and you know root, say x=a. Then you can write it as $$ \left( {x - a} \right)q\left( x \right) = 0 $$ and at by zero-product property you can try to solve the equation q(x)=0 which is of degree n-1. Of course if you have in the RHS a number different form zero this is not possible.