If the equation ax^2 +2bx−3c=0 has no real roots and (((3c)/4))< a+b, then


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Question Number 32382 by hizmzm1 last updated on 24/Mar/18

$If the equation {a x}^{2} + 2 b x - 3 c = 0 has$ $no real roots and (\frac{3 c}{4}) < a + b, then$
Commented byMJS last updated on 24/Mar/18

$I think {that}^{'} s not enough information$
Commented byMJS last updated on 24/Mar/18

$x^{2} + \frac{2 b}{a} x - \frac{3 c}{a} = 0$ $exactly one root :$ ${(x - r)}^{2} = x^{2} + \frac{2 b}{a} x - \frac{3 c}{a}$ $I . - 2 r = \frac{2 b}{a} \Rightarrow r^{2} = \frac{b^{2}}{a^{2}}$ $II . r^{2} = - \frac{3 c}{a}$ $\Rightarrow - \frac{3 c}{a} = \frac{b^{2}}{a^{2}} \Rightarrow c = - \frac{b^{2}}{3 a}$ $but now we have 1 root . It$ $disappears when we add a$ $real number > 0 \Rightarrow$ $\Rightarrow c > - \frac{b^{2}}{3 a}$ $\frac{3 c}{4} < a + b \Rightarrow c < \frac{4}{3} (a + b)$ $- \frac{b^{2}}{3 a} < c < \frac{4}{3} (a + b)$ $[\Rightarrow - \frac{b^{2}}{3 a} < \frac{4}{3} (a + b)$ $case 1 : a < 0$ $- b^{2} > 4 a (a + b)$ $- 4 a^{2} - 4 a b - b^{2} > 0$ $4 a^{2} + 4 a b + b^{2} < 0$ ${(2 a + b)}^{2} < 0 \Rightarrow no solution$ $case 2 : a > 0 leads to$ ${(2 a + b)}^{2} > 0$ $\Rightarrow a > 0]$
	Answered by Joel578 last updated on 24/Mar/18

	$no real roots \to D < 0$ ${(2 b)}^{2} - 4 a (- 3 c) < 0$ $4 b^{2} + 12 a c < 0$ $b^{2} > 3 a c$ $3 c < 4 (a + b)$ $3 a c < 4 a^{2} + 4 a b$ $3 a c < b^{2}$
	Commented byMJS last updated on 24/Mar/18

	$3 c < 4 (a + b) \Rightarrow 3 a c < 4 a^{2} + 4 a b with a > 0$ $3 c > 4 (a + b) \Rightarrow 3 a c > 4 a^{2} + 4 a b with a < 0$ $can you show a > 0 ?$
	Answered by saru53424@gmail.com last updated on 24/Mar/18

	$t h a n k s$
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