# What is the problem of getting 0 in the formula of zeros of the polynomial?

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#### Explanation

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A few thoughts...

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On the assumption that you are referring to getting

Given:

#ax^2+bx+c = 0" "# with#a!=0#

Multiply both sides by

#0 = 4a^2x^2+4abx+4ac#

#color(white)(0) = (2ax)^2+2(b)(2ax)+b^2-(b^2-4ac)#

#color(white)(0) = (2ax+b)^2-(b^2-4ac)#

Adding

#(2ax+b)^2 = b^2-4ac#

Take the square root of both sides, allowing for both

#2ax+b = +-sqrt(b^2-4ac)#

Subtract

#2ax = -b+-sqrt(b^2-4ac)#

Divide both sides by

#x = (-b+-sqrt(b^2-4ac))/(2a)#

This is known as the quadratic formula.

Now suppose

Immediately, we have an issue: The formula becomes:

#x = (-b+-sqrt(b^2))/0#

That is:

#x = (-2b)/0" "# or#" "x = 0/0#

Division by

If we go back up to the beginning of our derivation notice that we multiply through by

So it is really no surprise that the quadratic formula does not work when

Instead we have to treat

#x = -c/b#

...assuming

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