Problem with quadratic equations with parameters

[Bannini]

y=(p-1)x^2+(p+4)|x|+p+3
Find the function which has the zero x1=5

Alright, so I've been giving it a lot of thought and I haven't been able to solve this one.
There were a few similar problems with the given information of a minimum or maximum, but that was easy since it just followed
formula for alpha and beta of the extreme value. But now I've got a zero of a function and I'm not quite sure what to do with it.
I've tried using x1+x2=-b/a and x1x2=c/a and I get p=2/31 where I'm supposed to get p=2

I feel like I'm supposed to do something with the absolute value, but I don't know what. It has its use in the next part of the task where I have to
draw the graph, but other than that, I'm stumped.

 

[tkhunny]

1) Have you considered the Quadratic Formula?
2) You may need to split up that Absolute Value:

For x >= 0, we have y=(p-1)x^2+(p+4)x+(p+3)
For x < 0, we have y=(p-1)x^2-(p+4)x+(p+3)

 

[Bannini]

Here's what I get with my (probably flawed) logic

x;x>=0
-x;x<0
This is what we have for the absolute value
Matching that with x1=5 I would get
0=(p-1)x^2+(p+4)x+p+3
out of which I get:
31p-2=0
p=2/31

EDIT:
Ok, obviously, that doesn't work out since I need my p to be 2
If I take the second situation (-x,x<0);x1=5
I have:
(p-1)x^2-(p+4)x+p+3=0
I get 24p-48=0
p=2

Since I see now that it's the second situation that has the correct answer, could anyone explain why? I have no insight on why I should know that the correct answer is
in the second situation.

P.S. Thanks for the advice, tkhunny

 

[soroban]

Hello, Bannini!

Please check for typos.
As written, the problem does not make sense.

\(\displaystyle f(x) \:=\; (p-1)x^2+(p+4)|x|+p+3\)

Find the function which has the zero \(\displaystyle x_1=5.\)


I've tried using \(\displaystyle x_1+x_2=-\frac{b}{a}\) and \(\displaystyle x_1x_2=\frac{c}{a}\) and I get \(\displaystyle p =\frac{2}{31}\) . Yes!
. . where I'm supposed to get \(\displaystyle p=2.\) . This is wrong!

If \(\displaystyle p=2\), we have: .\(\displaystyle f(x) \:=\: x^2 + 6|x| + 5\)

. . And \(\displaystyle x=5\) is not a zero of this function.


Suppose \(\displaystyle x=5\) is a zero of \(\displaystyle f(x)\!:\;\;f(5) \,=\,0\)

We have: .\(\displaystyle (p-1)25 + (p+1)|5| + p+3 \:=\:0\)

. . . . . . . . \(\displaystyle 25p - 25 + 5p + 5 + p+3 \:=\:0\)

. . . . . . . . . . . . . . . \(\displaystyle 31p - 2 \:=\:0\)

. . . . . . . . . . . . . . . . . \(\displaystyle p \:=\:\frac{2}{31}\)

Your answer is correct!

 

[Bannini]

Ok now I'm totally lost
Here's the exact answer provided:
p=2; y=x^2-6|x|+5

It could be that they provided a wrong one or that there really is a typo.
The next step is to make the graph for that function and with p=2/31 it doesn't look all too well.

Either way, thanks for the help so far

 

[JeffM]

Ok now I'm totally lost
Here's the exact answer provided:
p=2; y=x^2-6|x|+5 Notice the minus 6 as the coefficient of |x|.

It could be that they provided a wrong one or that there really is a typo.
The next step is to make the graph for that function and with p=2/31 it doesn't look all too well.

Either way, thanks for the help so far

This is confusing.

y=(p-1)x^2+(p+4)|x|+p+3 Notice the plus (p + 4) as the coefficient of |x|.
Find the function which has the zero x1=5

\(\displaystyle p = 2 \implies (p - 1)x^2 + (p + 4)|x| + 5 = x^2\ PLUS\ 6|x| + 5 \ne x^2\ MINUS\ 6|x| + 5.\)

However, \(\displaystyle x = 5\ and\ y = x^2 - 6|x| + 5 \implies y = 5^2 - 6|5| + 5 = 25 - 30 + 5 = 0.\) 5 is a root for that function.

I suspect that either you misread the problem or the book misprinted the problem, which is supposed to be

\(\displaystyle y = (p - 1)x^2 - (p + 4)|x| + 5,\ and\ 5\ is\ a\ root.\ Find\ p.\)

\(\displaystyle 0 = (p - 1)5^2 - (p + 4)|5| + 5 \implies 25(p - 1) - 5(p + 4) + 5 = 0 \implies 25p - 25 - 5p - 20 + 5 = 0 \implies 20p - 40 = 0 \implies p = 2.\)

 

[Bannini]

Yeah, it's a misprint definitely, everything adds up nicely with that minus there.
Thanks for all the help people