Can someone help me with this question? I'm totally stuck.
The function f is defined by the expression f(x)= 2x+3 ∕ x − 2
For what values on a does the equation f(x) = a − x lack real roots?
I'm trying to figure out what values on a make no real roots. That's the premise, and I'm low on intuition.
>x^2 = ax - 2a -3
What values on a will make it so no real roots appear when I apply the pq-formula?
>The discriminant must be<0, so a^2-4(2a+3)<0
Why? Why can't it be >0?
How did you get this? All I have is x^2 = ax - 2a - 3, and I don't understand how you came to a^2-4(2a+3)<0. I hope me requesting explanations isn't too tedious.
Since my school doesn't have math teachers I haven't had most of algebra explained/ taught to me. We only get a text book to work through and there is no info on how to solve equations like "x^2 = ax - 2a - 3" in it.
> op fuck you. it's like 3 am here
2x+3 ∕ x − 2 =a - x
x should be different than 2;(because it's division by 0);
2x+3 = ( a - x ) * (x-2) <=>
ax-2a -x^2 +2x -2x -3=0 <=>
-x^2 + ax +2x -2x -2a -3 =0 <=>
-x^2 + ax -2a -3 =0 ;
we need to find the roots where -x^2 =0;
thus delta is
a^2 -4(-1 *(-2a-3)) <=> a^2 - 8a +12 .
for x to not have real we need delta <0
thus a^2 -8a +12 must be < 0.
a^2 -8a +12 =0 <=> delta 2 is
(-8)^2 -4(1*12) = 64 - 48 = 16.
the lowest point is -b/(2a') = 8/2 =4. because a is always positive a^2 -8a +12 cannot be negative... and x will always have solutions. i might have made bad calculations but in theory that is what you have to do.
suck my dick i love math.
i think you don't understand some basic algebra here.
imagine the x^2 curve. if you cant, juste type in x^2 on google, you'll get one. all other functions that are written like : ax^2 + bx + c are variants of it, slimmer, larger, turned upside down or not, with their rounded corner going either up or down, left or right in the XY axis, depending on what values a, b, and c are.
Now, the question is, the delta indicates if your x^2 type parabole touches the x-axis line, you can see on wikipedia why, but in delta = b^2 - 4ac , following what i said about how a,b,c influes on the position of your function on the axis, you can easily understand that delta may tell you where your function may be located.
Delta > 0 means it touches the X axis with two branches, therefore you got two ways to touch the x axis with your functions, i.e two solutions
delta = 0 means you only touch the x= 0 axis (or x axis) with the "butt" of the parabole, that is the rounded corner.
delta<0 means the function does not touch the X axis, therefore there is no way oh having f(x) = 0, i.e no solutions.
As for the answer to your questions, i'm pretty sure >>46356 gives you the right answer.
It's simple, you've already solved most of the problem yourself anon.
You worked out the equation x^2+2a+3-ax=0
Rearrange it as x^2-ax+2a+3 = 0
Now, for exposition ourposes, I'm gonna call a here, say, p.
So let's rewrite it as x^2-px+2p+3
This is a quadratic equation, remember that a quadratic equation is one written in the form ax^2+bx+c, in this case a = 1, b =-p and c=2p+3
Now, the solution to a quadratic equation is given by the quadratic formula, (-b ± sqrt(b^2-4ac))/2a
So, for the values in this equation:
(p ± sqrt(p^2-4(2p + 3))/2
Now, this equation will clearly have a real solution for all values where the square root has a real solution, that is, where p^2-4(2p + 3) >= 0, so it will not have one where p^2-4(2p + 3) < 0
Simplifying the expression we get p^2 -8p -12 < 0
Now, remember that p stood for a in the original question, where it asked to solve find where f(x)= a -x had no real roots.
So, a^2-8a -12 < 0
This gives us two solutions, a< 4 - 2*sqrt(7) and a<4 +2*sqrt(7), or just a <4+2*sqrt(7), as the inequality holds at the middle point of the interval between those two values, a=4.
>>a^2 -4(-1 *(-2a-3)) <=> a^2 - 8a +12
The 12 there is negative, everything else that follows is wrong.
Anon, >>46356 made a small mistake and the solution isn't correct, I corrected it in >>46458
If you really a need a one line answer, it would be a <4+2*sqrt(7) but I'd strongly encourage you to study the answers you've received so that you can see the reasoning and be able to solve similar problems in the future.
Okay, but if a = 2, then x = 2 ± sqrt 2^2 - 11 is the same as x = 2 ± sqrt (-7), which doesn't work, or something... shouldn't the answer be let's say n>a or n<a... what should replace n then?
I fucking hate my school for not having enough math teachers. Our whole fucking country is in lack of math teachers because all we have is goddamn gender studies and quasi-sociologists.
Sorry, made a small mistake checking the range, the solution is 4-2*sqrt(7)<a<2+sqrt(7)
Anyway, the key is to practice a lot, and, really, don't rely on teachers, despite what the common word is.Get yourself a good math book at the level you're studying and study from that. This is an elementary algebra problem, some books I could recommend for at this level are "Algebra: a very short introduction" by Peter M. Higgins and "Algebra: A Complete Introduction" by Hugh Neill.
I will check it out. I appreciate you sharing your wisdom. I acknowledge the fact that I have no one to blame but myself for not knowing what I should know, in these times when everything is 'out there'.
See one thing I don't get, and it feels like I'm following you for the most part, is where a< 4 - 2*sqrt(7) comes from?
I have a^2-8a -12, and through the quadratic formula does that not become: a = 4 ± sqrt (4^2+12=28)..? I'm not following.