Question

The question

Answer 1

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.

Answer 2

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

Answer 3

ya this one

Answer 4

And they have to be real number solutions, correct?

Answer 5

the x solutions i mean

Answer 6

yes

Answer 7

\(\large\color{black}{ \displaystyle x^2-2x+p^2-3p-4=0 }\)

\(\large\color{black}{ \displaystyle x^2-2x=-p^2+3p+4 }\)

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

Answer 8

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...

Answer 9

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

Answer 10

\(\large\color{black}{ \displaystyle -p^2+3p+5 }\)

\(\large\color{black}{ \displaystyle -(p^2-3p)+5 }\)

\(\large\color{black}{ \displaystyle -(p^2-3p)-\frac{9}{4}+\frac{9}{4}+5 }\)

\(\large\color{black}{ \displaystyle -\left(p^2-3p+\frac{9}{4}\right)+\frac{9}{4}+5 }\)

\(\large\color{black}{ \displaystyle -\left(p^2-3p+\frac{9}{4}\right)+\frac{29}{4} }\)

\(\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2+\frac{29}{4} }\)

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

Answer 11

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

Answer 12

excellent

Answer 13

the solutions seems little longer

Answer 14

\[\large \color{black}{\begin{align} x^2-2x+\color{blue}{p^2-3p-4}=0 \hspace{.33em}\\~\\
\end{align}}\]
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 :
\[\color{blue}{p^2-3p-4}\lt0\\~\\\color{blue}{(p+1)(p-4)}\lt 0\\~\\ \color{blue}{-1\lt p\lt 4}\]

Answer 15

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

Answer 16

lol xD

Answer 17

Perfect!