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I know that a quadratic equation can be represented in the form $$ax^2 + bx + c = 0$$ where $a$ is not equal to $0$, and $a$, $b$, and $c$ are real numbers. However, if there is an equation in the form $$ax^2 + bx = 0$$ would it be classified as a quadratic equation since the conditions are satisfied, or would it be a linear equation since it can be simplified into $ax + b = 0$?

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  • $\begingroup$ We define a polynomial by the highest power present $\endgroup$ Apr 19, 2021 at 8:30
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    $\begingroup$ It can't be simplified into $ax+b=0$, which has one solution. It can be factorised, like other quadratics, as $x(ax+b)=0$; and this gives two solutions, in the normal way. $\endgroup$
    – TonyK
    Apr 19, 2021 at 8:50

4 Answers 4

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It is a quadratic equation as it satisfies the definition.

Notice that $ax^2+bx=0$ and $ax+b=0$ are not equivalent, the first one has $0$ as a solution for sure and $\frac{-b}a$ as a root as well.

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  • $\begingroup$ Oh I see. Thanks! $\endgroup$
    – Twilight
    Apr 19, 2021 at 8:31
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A quadratic equation is an equation that can be rearranged as $ax^2+bx+c=0$ where $a$ is not equal to $0$ and $b$ and $c$ are real numbers. If $a=0$ then the equation is linear not quadratic since the $x^2$ has no influence .

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Hints

  1. If you draw the graph of $y=ax^2+bx$ what shape is it? (Plug in some non-zero values for $a$ and $b$)

  2. If you factorize $ax^2+bx=0$ and then apply null factor law, how many solutions are there?
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You need only $a≠0$ for $$ax^2+bx+c=0.$$

If the degree of your polynomial is equal to $2$, then you have a quadratic polynomial. This means , your equation $ax^2+bx=0$ is still a quadratic equation, if $a≠0.$

But, the degree of the polynomial $ax+b$ is equal to $1$. This implies, $ax+b=0$ is not a quadratic.


Small Supplement:

$ax^2+bx=0$ is not equivalent to $ax+b=0$. Because, $x=0$ is not always a root of $ax+b=0.$

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  • $\begingroup$ I see. But upon multiplying x to both sides of ax + b = 0, we get ax^2 + bx = 0. Would this be a quadratic equation (since a is not equal to 0)? Or would it be linear since it is the same as ax + b = 0? $\endgroup$
    – Twilight
    Apr 19, 2021 at 8:30
  • $\begingroup$ @Twilight I added a small supplement. $\endgroup$ Apr 19, 2021 at 8:41
  • $\begingroup$ @Twilight: Note that you cannot multiply by something on both sides of an equation until you're sure that that something is not zero. $\endgroup$ Apr 19, 2021 at 8:44

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