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Very basic stuff from school we know that we can calculate zero places of quadratic function which has form $ax^2 + bx + c$ and we assume that $a \neq 0$, now what if $a=0$? Why can't we use delta to calculate zero places of linear function? I know it's very easy to calculate zero place of linear function, but i'm quite interested in such stuff. Thanks for any response.

EDIT : Right, to make my question more understandable: From where such formula of roots of quadratic function come from? And as interesting as possible, elaborate this topic! i'll be greatly grateful.

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3 Answers 3

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Clearly, using the quadratic formula to find the "zeros" (roots) of a quadratic: solving for $ax^2 + bx + c = 0$, is not defined for $a = 0$, since the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ is undefined when $a = 0$: we cannot be divided by $2a = 2\cdot 0 = 0$. And furthermore, when $a = 0$, the equation is no longer quadratic; it defines a line.

So, when $a$ in $ax^2 + bx + c = 0$, and $a = 0$, we have $bx + c = 0$, and provided $b$ is not zero, then the single zero (solution for $x$) is given by $$bx = -c \implies x = -\dfrac cb, \quad b\neq 0.$$


EDIT: You might want to read about the Quadratic Equation in Wikipedia. The entry covers the history behind the formula, explains how to derive the formula and why it gives the roots (zeros) of a quadratic, lists variations on/alternatives to the formula, and alternative methods for solving such equations, and ways in which it can be used/adapted for higher dimensional equations.

See also Quadratic formula @ sosmath.com

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Certainly deserves a thumbs up! +1 –  Amzoti May 11 '13 at 0:33
    
+1 $-------{---(@$ –  Babak S. Jun 11 '13 at 4:54
    
That's really sweet, @Babak! –  amWhy Jun 11 '13 at 4:56

I don't know what you mean by "use delta," but the zero of $bx+c$ is given by $x=-\frac{c}{b}$, which you can check by plugging in.

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ah, okay sorry. delta is given by $delta = b^2 - 4ac$ and zero places of quadratic function is $x1=\frac{-b+\sqrt(delta)}{2a}$ and $x2=\frac{-b-\sqrt(delta)}{2a}$. –  Chris Jan 8 '13 at 19:47

Let us fix $b$ and $c$, with $b\gt 0$. Think of $a$ as variable.

You are undoubtedly familiar with the fact that if $a\ne 0$, then the roots of the quadratic $ax^2+bx+c=0$ are given by $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$ The above formula is troublesome at $a=0$, for obvious reasons!

We sidestep some unpleasantries by looking only at the root $\frac{-b+\sqrt{b^2-4ac}}{2a}$. We will examine the behaviour of this root as $a$ approaches $0$.

First note that if $a\ne 0$, then $$x=\frac{-b+ \sqrt{b^2-4ac}}{2a}=\frac{-b+ \sqrt{b^2-4ac}}{2a}\cdot \frac{b+\sqrt{b^2-4ac}}{b+\sqrt{b^2-4ac}}.$$ But $\left(-b+\sqrt{b^2-4ac}\right)\left(b+\sqrt{b^2-4ac}\right)=-4ac$, so we obtain $$x=\frac{-4ac}{(2a)\left(b+\sqrt{b^2-4ac}\right)}=\frac{-2c}{b+\sqrt{b^2-4ac}}.$$ Now we can safely let $a$ approach $0$. As $a$ approaches $0$, $b+\sqrt{b^2-4ac}$ approaches $2b$. It follows that the root of the quadratic approaches $\frac{-2c}{2b}$, which is the right answer for the root of the linear equation $bx+c=0$.

Remark: Much too elaborate to be an answer to your question! But I thought it would be interesting to show how the solution of the linear equation can be viewed as a limit of the standard solution of the quadratic.

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For those who might be interested, this used to be a standard topic in quadratic equations (especially in the 1800s), at least in texts written in English. See the google-books search for the simultaneous search of the word "quadratic" and the phrase "infinite root". –  Dave L. Renfro Jan 9 '13 at 15:32

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