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Discriminant

If the equati...

Question

If the equation $a x^{2} + b x + c = 0$ has non-real roots,prove that $1 + \frac{c}{a} + \frac{b}{a} > 0$

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Solution

The equation $a x^{2} + b x + c = 0$ is non-real
$\Rightarrow b^{2} - 4 a c < 0$
$\Rightarrow b^{2} < 4 a c$
$\Rightarrow | b | < 2 \sqrt{a c}$
$\Rightarrow b < a + c$
$\Rightarrow a + c - b > 0$
$\Rightarrow a + b + c > 0$
As A.M $>$ G.M
$\Rightarrow \frac{a + c}{2} > \sqrt{a c}$
$\Rightarrow a + c > 2 \sqrt{a c}$
$\Rightarrow \frac{a + c}{2} > \sqrt{a c}$
$\Rightarrow - b < \sqrt{a c}$
$\Rightarrow a + b + c > 0$
$\Rightarrow \frac{a}{a} + \frac{b}{a} + \frac{c}{a} > 0$ by dividing by $a$
$\Rightarrow 1 + \frac{b}{a} + \frac{c}{a} > 0$
Hence proved

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