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

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Feb 14, 2018

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On the assumption that you are referring to getting #a=0# in the quadratic formula, let us see what it is all about.

Given:

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

Multiply both sides by #4a# to get:

#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 #(b^2-4ac)# to both ends, we find:

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

Take the square root of both sides, allowing for both #+# and #-# square roots to find:

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

Subtract #b# from both sides to get:

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

Divide both sides by #2a# to get:

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

This is known as the quadratic formula.

Now suppose #a=0#.

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 #0# is undefined, so the quadratic formula fails to give us the correct answer.

If we go back up to the beginning of our derivation notice that we multiply through by #4a#. If #a=0# then we just multiplied the whole equation by #0#, resulting in an equation that will give us no information.

So it is really no surprise that the quadratic formula does not work when #a=0#.

Instead we have to treat #a=0# as a separate case, in which the polynomial should no longer be considered a quadratic and the solution is:

#x = -c/b#

...assuming #b != 0# of course.

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