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If $a,b,c \in R$ such that $abc \neq0$ If $x_1$ is a root of $a^2x^2+bx+c=0, x_2$ is a root of $a^2x^2-bx-c=0 $ and $x_1 > x_2 >0$ then the equation $a^2x^2+2bx+2c=0$ has roots $x_3$ .

Prove that $x_3$ lies between $x_1 \& x_2$

Let f(x) = $a^2x^2+2bx+2c=0$

$\Rightarrow f(x_1)=a^2x_1^2+2bx_1+2c=-a^2x_1^2$

$\Rightarrow f(x_2) = a^2x_2^2+2bx_2+2c=3a^2x_2^2$

$\Rightarrow f(x_1)(x_2) = (3a^2x_2^2)(-a^2x_1^2) <0$

$\Rightarrow $Thus one root of $a^2x^2+2bx+2c=0$ will lie between $x_1 \& x_2$

Please provide explanation on the last statement of this answer how it derived ...Thanks..

Sachin
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    Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. – Lord_Farin Apr 27 '13 at 15:46
  • The recent edit has made this question worth reopening (@Lord_Farin) – The Chaz 2.0 Apr 27 '13 at 17:46
  • Thanks for adding further context! – Lord_Farin Apr 27 '13 at 21:30

2 Answers2

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It's a consequence of the intermediate value theorem, roughly:

For a continous function $f$ on interval $[a,b]$ of $\mathbb{R}$, if $f(a)f(b)<0$ then there exists $c\in(a,b)$ such that $f(c)=0$

ioveri
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Make a rough sketch of your findings. Will the curve cut the x-axis?

Mick
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