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

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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Feb 14, 2018

A few thoughts...

#### Explanation:

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

Given:

$a {x}^{2} + b x + c = 0 \text{ }$ with $a \ne 0$

Multiply both sides by $4 a$ to get:

$0 = 4 {a}^{2} {x}^{2} + 4 a b x + 4 a c$

$\textcolor{w h i t e}{0} = {\left(2 a x\right)}^{2} + 2 \left(b\right) \left(2 a x\right) + {b}^{2} - \left({b}^{2} - 4 a c\right)$

$\textcolor{w h i t e}{0} = {\left(2 a x + b\right)}^{2} - \left({b}^{2} - 4 a c\right)$

Adding $\left({b}^{2} - 4 a c\right)$ to both ends, we find:

${\left(2 a x + b\right)}^{2} = {b}^{2} - 4 a c$

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

$2 a x + b = \pm \sqrt{{b}^{2} - 4 a c}$

Subtract $b$ from both sides to get:

$2 a x = - b \pm \sqrt{{b}^{2} - 4 a c}$

Divide both sides by $2 a$ to get:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

This is known as the quadratic formula.

Now suppose $a = 0$.

Immediately, we have an issue: The formula becomes:

$x = \frac{- b \pm \sqrt{{b}^{2}}}{0}$

That is:

$x = \frac{- 2 b}{0} \text{ }$ or $\text{ } x = \frac{0}{0}$

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 $4 a$. 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:

$x = - \frac{c}{b}$

...assuming $b \ne 0$ of course.

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