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Given equation: $$x^2 + 5x + c = 0$$ For what $c$'s will the equation have non-real solutions?

I have try to plug in all kinds of different values for $c$ and then solve the equation to see if it gives me complex solution.

My question: Is there a theorem/lemma or a formula I need to state to answer this question?

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  • $\begingroup$ Hint: Just use the Quadratic Formula and analyze that result for the roots. $\endgroup$ – Moo Sep 16 '16 at 1:26
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    $\begingroup$ Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases. $\endgroup$ – Semiclassical Sep 16 '16 at 1:29
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You have exactly one root if you have a perfect square.

Compete the square:

$(x^2 + 2 \frac 52 x + \frac {25}{4}) = (x+\frac 52)^2$

If $c = \frac {25}{4}$ there is one (real) root.

If $c >\frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.

if $c > \frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.

Alternate: apply the quadratic formula

If you know the quadratic formula, then what does it take to get a negative under the radical?

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  • $\begingroup$ Great explanation. I was not thinking about square root or graphing it. Thanks! $\endgroup$ – Cesar Agama Sep 16 '16 at 6:19
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Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$

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The discriminant formula for any quadratic is-
$D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.

  • If $D>0$, then the quadratic has two real and distinct roots.
  • If $D=0$, then the quadratic has two real and equal roots.
  • If $D<0$, then the quadratic has two complex roots.

In your question-
$$a=1, b=5, c=?$$ $$D=5^2 - 4×1×c =25 - 4c$$

The equation has non-real roots only if-
$D<0$
$\Rightarrow 25 - 4c <0$
$\Rightarrow c < 25/4$ $_{(Answer)}$

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