## mathmath333 one year ago The question

1. mathmath333

Find the range of $$'p\ '$$ if roots of the equation \large \color{black}{\begin{align} x^2-2x+p^2-3p-4=0 \hspace{.33em}\\~\\ \end{align}} are opposite in sign.

2. SolomonZelman

what do you mean that the roots are opposite in sign, one x solution is negative and the other x solution is positive ?

3. mathmath333

ya this one

4. SolomonZelman

And they have to be real number solutions, correct?

5. SolomonZelman

the x solutions i mean

6. mathmath333

yes

7. SolomonZelman

$$\large\color{black}{ \displaystyle x^2-2x+p^2-3p-4=0 }$$ $$\large\color{black}{ \displaystyle x^2-2x=-p^2+3p+4 }$$ I am going to treat $$\large\color{black}{ \displaystyle -p^2+3p+4 }$$ $$\large\color{black}{ \displaystyle x^2-2x+1=-p^2+3p+4+1 }$$ making a perfect square trinomial and factoring $$\large\color{black}{ \displaystyle (x-1)^2=-p^2+3p+5 }$$ $$\large\color{black}{ \displaystyle x-1=\pm\sqrt{-p^2+3p+5} }$$ $$\large\color{black}{ \displaystyle x=1\pm\sqrt{-p^2+3p+5} }$$ this is what I am up to so far

8. SolomonZelman

now you need to find the p-range that will make one pos. and one neg. solut. for x. I will think how to do it percisely, don't want to make an err anywhere...

9. SolomonZelman

oh, the 3rd line should say I am going to treat "-p^2+3p+4 as a constant

10. SolomonZelman

$$\large\color{black}{ \displaystyle -p^2+3p+5 }$$ $$\large\color{black}{ \displaystyle -(p^2-3p)+5 }$$ $$\large\color{black}{ \displaystyle -(p^2-3p)-\frac{9}{4}+\frac{9}{4}+5 }$$ $$\large\color{black}{ \displaystyle -\left(p^2-3p+\frac{9}{4}\right)+\frac{9}{4}+5 }$$ $$\large\color{black}{ \displaystyle -\left(p^2-3p+\frac{9}{4}\right)+\frac{29}{4} }$$ $$\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2+\frac{29}{4} }$$ now, this (the whole expression) has to be greater than 1, so that x is positive or negative. $$\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2+\frac{29}{4}>1 }$$ $$\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2>1-\frac{29}{4} }$$ $$\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2>-\frac{25}{4} }$$ $$\large\color{black}{ \displaystyle \left(p-\frac{3}{2}\right)^2<\frac{25}{4} }$$ $$\large\color{black}{ \displaystyle \left(p-\frac{3}{2}\right)^2<\frac{25}{4} }$$ so this p solution is -1<p<4

11. SolomonZelman

sorry for long posting. if you prefer a verbal expln. pliz let me know, i will try my best.

12. mathmath333

excellent

13. mathmath333

the solutions seems little longer

14. ganeshie8

\large \color{black}{\begin{align} x^2-2x+\color{blue}{p^2-3p-4}=0 \hspace{.33em}\\~\\ \end{align}} The constant term here represents the product of roots, so it must be negative for the quadratic equation to have roots that are of different signs : $\color{blue}{p^2-3p-4}\lt0\\~\\\color{blue}{(p+1)(p-4)}\lt 0\\~\\ \color{blue}{-1\lt p\lt 4}$

15. Loser66

@ganeshie8 I have to log in to say: " You are a math wicked"

16. ganeshie8

lol xD

17. mathmath333

Perfect!