We have a quadratic equation $ax^2$+ $bx$ + c =0 . It has no real roots. i.e $b^2$- $4ac$ < 0 .What is the sign of c ?

My tries- I have supposed the $4ac$ will be greater than $b^2$ , because if it wasn’t then the roots will never be less then zero . So Two cases arise-

1- Both $a$ and $c$ are greater then zero.

2- Both $a$ and $c$ are smaller then zero.

So $c$ could be greater than zero or smaller than zero .

But my book says, $c<0$ . WHY?

  • 1
    $\begingroup$ Your book, if you are copying it correctly, is wrong. $\endgroup$ – mweiss Oct 26 '15 at 14:09
  • $\begingroup$ Is there any implicit assumption they made about the quadratic equation? $\endgroup$ – Element118 Oct 26 '15 at 14:10
  • $\begingroup$ Have you been told anything specific at all about $a$ or $b$? Because $x^2 + x + 1$ has no roots and neither does $-x^2 - x - 1$. So unless you know something more (like, for instance, it says monic polynomial), then $c$ can be anything. $\endgroup$ – Arthur Oct 26 '15 at 14:10
  • $\begingroup$ The value of $c$ will depend upon $a$ and $b$ $\endgroup$ – user210387 Oct 26 '15 at 14:11
  • $\begingroup$ Your equation is $(a+b)x^2=-c$ Are you sure? $\endgroup$ – Konstantinos Gaitanas Oct 26 '15 at 14:12

As you have correctly noted in the original post, what matters is not the sign of $c$, but rather the sign of $b^2-4ac$. Here are a few possibilities:

  1. If $a$ and $c$ have opposite signs, then $4ac$ is negative, and hence $b^2-4ac$ will be positive, so the equation will have two real solutions.
  2. Therefore if the equation has no real solutions, then $a$ and $c$ must have the same sign.
  3. However, while the condition that $a$ and $c$ have the same sign is necessary, it is not sufficient to ensure that there are no real solutions. In order to ensure that there are no solutions, $a$ and $c$ have to have the same sign and have a product that is larger than $b^2$.
  • $\begingroup$ May I ask why the downvote? $\endgroup$ – mweiss Oct 26 '15 at 18:33

Two cases :
Your equation is $(a+b)x^2=-c$.(I suppose you have a typo ).This does not tell us anything about the sign of $c$.
Your equation is $ax^2+bx+c=0$ .This does not tell us anything about the sign of $c$.
But what you can say is that the sign of $c$ is the same with $a$ in order to have $D<0$ .

  • $\begingroup$ Yes sure! It does not tell us anything about c . But if D<0 4ac must be greater than b^2 $\endgroup$ – Aaryan Dewan Oct 26 '15 at 14:59

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