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$a$ , $b$, $c$ are real numbers where a is not equal to zero and the quadratic equation \begin{align} ax^2 + bx +c =0 \end{align} has no real roots then prove that $c(a+ b+ c)>0$ and $a(a+ b + c) >0$

My Approach : As the equation has no real roots then its discriminant is less than zero. So the graph of the equation will be above $x$-axis or below $x$-axis . I am able to conclude signs of $a$ , $b$, $c$ but still not getting appropriate answer. Please explain the concept......

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Continuing from where you left.

Let $p(x)=ax^2+bx+c$

Observe that

  • $p(1)=(a+b+c)$
  • $p(0)=c$
  • $\lim_{x \rightarrow \infty}\frac{p(x)}{x^2} =a$

So, $$p(1) \cdot p(0)=c(a+b+c) >0$$ $[p(0)$ and $p(1)$ have same sign]

Also, $$p(1) \cdot \lim_{x\rightarrow \infty} \frac{p(x)}{x^2} =a(a+b+c) >0$$ [$p(1)$ and $\lim_{x \rightarrow \infty} \frac{p(x)}{x^2}$ have the same sign because $x^2 > 0$, for all $x \in \mathbb{N}$]

Hence, the claim is proved.

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  • $\begingroup$ Whay: $\lim_{x \rightarrow \infty}p(x)=a$ .?? $\endgroup$ – Emilio Novati Jun 28 '16 at 15:02
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    $\begingroup$ Perhaps you meant $\lim_{|x| \to \infty} \frac{p(x)}{x^2}=a$, thus for large enough $x$, $p(x)$ has the same sign as $a$. $\endgroup$ – Ian Jun 28 '16 at 15:03
  • $\begingroup$ Prove your third bullet $\endgroup$ – Aman Rajput Jun 28 '16 at 15:03
  • $\begingroup$ @Ian Thanks. Now, fixed. $\endgroup$ – Dragonemperor42 Jun 28 '16 at 15:07
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For the second part, another justification is that since the discriminant is less than zero, then it must hold that $a,c$ have the same sign. Why? If $a,c$ had opposite signs then the discriminant $\Delta = b^2-4ac$ would have been positive.

Since $c,(a+b+c)$ have the same sign and $c,a$ have the same sign then $a,(a+b+c)$ have the same sign as well, thus: $$a(a+b+c) >0.$$

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If the quadratic equation has no real roots, then the graph of the corresponding quadratic function ($y = ax^2 + bx + c$) will not cross the x-axis. (Case-1 I will just discuss the case that the function states in the upper part of the x-axis.) Then,

(1) the parabola must be opening upward. This means $a > 0$;

(2) it will somehow cut the y-axis at $(0, c)$ where $c > 0$; and

(3) for all real t, $y(t) = at^2 + bt + c >0$. This is especially true for $t = 1$. That is $y(1) = … = a + b + c > 0$.

Required results for this case follow.

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