quadratics inequations

[outer belfry]

i got -7 < x > 5 where there is no real roots however the answer is -7 < x < 5 and i dont get why

[lyric gazelle]

So using discriminant we have
P^2 + 2p + 1 -36 < 0
p^2 + 2p - 35 < 0
(p+7)(p-5) < 0
So critical values are x=-7 and 5
If you draw this quadratic and see what’s below y=0 you’ll see that it’s the part between the x values of -7 and 5 so the answer is -7<x<5

[outer belfry]

i get all of that

except how x is less than 5

[wooden mulch]

draw the function

[lyric gazelle]

Did you draw the graph

[wooden mulch]

on a graph

and find the range of values for which y<0

its going to be between -7 and 5

therefore -7<x<5

[outer belfry]

yeah

im trying to send it hollip

[lyric gazelle]

Ok

[outer belfry]

you can see that when it touches 5 it goes under the graph so how does that make it not a real root

but after 5 it goes above the line

therefore its not a real root after 5?

[lyric gazelle]

Cuz this isn’t the original graph it’s the graph of the discriminant

[outer belfry]

so whats the original graph

[lyric gazelle]

All we are trying to do is solve the quadratic we have your over complicating it

[outer belfry]

hm

[lyric gazelle]

You don’t know

[outer belfry]

ok so then how do i know that its not a real root when its under 5?

[lyric gazelle]

Because that’s the solution to the equation we got from the discriminant

[outer belfry]

i see

so if its below y =0 its when it has no real root

and then after -7 the same?

[lyric gazelle]

Yeah

[wooden mulch]

its this region

so its between -7 and 5

-7<x<5

[outer belfry]

ohhh

i completely misunderstood the greater and less signs there

[lyric gazelle]

The solution tells us that when p is between -7 and 5 there is no real solutions to the original quadratic

[outer belfry]

i get it

that makes a lot of sense

i thoguht -7< means -8,-9 lol