If-the-equation-ax-2-2bx-3c-0-has-no-real-roots-and-3c-4-lt-a-b-then-

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 by MJS last updated on 24/Mar/18

I think {that}^{'} s not enough information

Commented by MJS 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 by MJS 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