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Find the range of p for which one root of the equation

$$0= x^2 - (p+1)x + (p^2 + p-8)$$

is greater than 2 and of the other root is smaller than 2.

How i can achieve this? Thanks in advance.

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1  
Should $y$ be $0$ or do you have to solve the problem given any value of $y$? –  Rankeya Mar 17 '12 at 5:21
    
I have started to write an answer putting $y=0$. sigh Thanks for asking it! :-) –  user21436 Mar 17 '12 at 5:22
    
THanks @Rankeya. i have edited my question. –  vikiiii Mar 17 '12 at 5:25

4 Answers 4

up vote 1 down vote accepted

Hint:

By root, I assume you mean find $x$ such that $y = 0.$ Well, if $$ 0 = x^2 - (p+1)x + (p^2 + p-8) $$ then the roots $r_0, r_1 $ are given by: $$ r_0 = \frac{p+1 + \sqrt{(p+1)^2 - 4 (p^2 +p -8)}}{2} \\ r_1 = \frac{p+1- \sqrt{(p+1)^2 - 4 (p^2 +p -8)}}{2} $$ To satisfy the conditions, set $r_0 > 2, r_1 <2 .$ So $$ p+1 + \sqrt{33-2 p-3 p^2} > 4 \\ p+1 \color{red}{-} \sqrt{33-2 p-3 p^2} < 4$$ Try to solve for $p$ from here.

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1  
Dear @J.D.: I think the last line should have a "-" sign after $p+1$. –  Rankeya Mar 17 '12 at 5:27
    
@J.d - After that i will get 2 ranges from different equation.I need to take intersection of that..right? –  vikiiii Mar 17 '12 at 5:28
    
Yes, precisely. –  Rankeya Mar 17 '12 at 5:30
    
@Rankeya thank you. –  user2468 Mar 17 '12 at 5:32

Now, you know that the roots of your equation are $2+a$ and $2-b$ for $a,b \gt 0$. Now, try and transform this into an equation in $a$ and $-b$. This is easy to do.

Convince yourself that this transformation is $y=x+2$.

Now, the only condition you have is that the product of the roots is negative.

So, you should end up with, $$\begin{align}p^2+p-8-2(p+1)+4 \lt 0\\p^2-p-6 \lt 0\\(p+2)(p-3) \lt 0\\p \in(-2,3)\end{align}$$

As André points out, there is no need to check whether the roots are real, because if they were complex, since complex roots occur in conjugate pairs we'll have their product is positive. (Thanks are due to Andre pointing this out. This makes this solution even more shorter and perfectly right.) $$\boxed{p \in (-2,3)}$$

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@Kannapan Sampath: There is no need to check whether the roots are real. For if they were non-real, they would be a pair of complex conjugates. But then their product would be positive. –  André Nicolas Mar 17 '12 at 6:04
    
Ah, Thank You, I missed on this. But, it does not affect here. :-) –  user21436 Mar 17 '12 at 6:07
    
@AndréNicolas I have written that out. How does it look? –  user21436 Mar 17 '12 at 6:20
    
@Kannapan Sampath: Looks fine. –  André Nicolas Mar 17 '12 at 6:27

Since the coefficient of $x^2$ is positive, the equation represents a upward opening parabola,

For one root of the equation to be greater than 2 the other root to be smaller than 2,the value of quadratic must be negative at 2.

y(2) < 0

$$2^2 - (p+1)2 + (p^2 + p-8) <0$$

$$ p^2-p-6 <0$$

$$ (p+2)(p-3) <0$$

$$\boxed{p \in (-2,3)}$$

But for roots to exist the discriminant of the equation $$y= x^2 - (p+1)x + (p^2 + p-8)$$

must be positive,

$$(p+1)^2 - 4(p^2 + p-8) >0$$ $$-3p^2-2p+33>0$$ $$(3p+11)(3-p)>0$$ $$(3p+11)(p-3)<0$$

$$\boxed{p \in (\frac{-11}{3},3)}$$

taking union of inequalities (1) and (2) we get

$$\boxed{p \in (-2,3)}$$

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1  
BTW: $x \tag{1}$ will give $x \tag{1}$ –  user2468 Mar 17 '12 at 5:38
    
But, if I use this piece in n(th) line, then x comes in the n + 1 (next line), what should we do so as to bring x one line upwards. –  Tomarinator Mar 17 '12 at 14:25
    
Good question! I have no idea :) –  user2468 Mar 17 '12 at 15:36

A parabola $f(x)$ with a positive coefficient of $x^2$ has two roots such that one is lower than $a$ and the other is greater than $b>a$ iff $f(a)<0$ and $f(b)<0$.

This guarantees both -- the existence of two roots, and the condition above.

In your case, $a=b=2$. So, you get only one equation: $f(2)=p^2-p-6<0$, and the range for $p$: $(-2,3)$.

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