Determine where solutions exist

Hello! I'm having a little difficulty practicing for the upcoming Mat tests. On questions such as

x⁴ = (x-c)^2, has four real solutions for

c <= 1/4, -1/4 <= c <= 1/4 (I think!), c <= -1/4, all values of c

I don't know what method to use, or really how to even start.

I see that there cant be a real solution if (x-c)^2 <= 0 but I don't know where to go next. There are lots of these sorts of questions, almost 4 or 5 per paper so I assume I must be missing something big, or there is some method(s) to try

thanks in advance

Reply 1

Rainingshame

13

Original post by Lemmingz95

Hello! I'm having a little difficulty practicing for the upcoming Mat tests. On questions such as

x⁴ = (x-c)^2, has four real solutions for

c <= 1/4, -1/4 <= c <= 1/4 (I think!), c <= -1/4, all values of c

I don't know what method to use, or really how to even start.

I see that there cant be a real solution if (x-c)^2 <= 0 but I don't know where to go next. There are lots of these sorts of questions, almost 4 or 5 per paper so I assume I must be missing something big, or there is some method(s) to try

thanks in advance

Well a simpler way to solve it might be to square root either side, though in doing so you have to consider the situation for both + and -. You'll then get two quadratics and you should surely know how to find out if a quadratic has real roots or not.

Reply 2

Lemmingz95

OP

Sorry if I wasn't very clear. here's the full question

Given a real constant c, the equation:
x⁴= (x - c)^2
has four real solutions (including possible repeated roots) for

(a) c<= .25, (b) -.25 <= c <= .25, (c) c <= -.25, (d) all values of c

Reply 3

Lemmingz95

OP

and yes, I really, really should have square rooted!

as its given that there exists 4 solutions, you know that b^2 - 4ac has to be greater than 0 ... and it's simple from there

Reply 4

TenOfThem

18

Original post by Lemmingz95

and yes, I really, really should have square rooted!

as its given that there exists 4 solutions, you know that b^2 - 4ac has to be greater than 0 ... and it's simple from there

I would not square root per se

I would use difference of 2 squares

It has the same outcome but is neater IMO

(x^2 + (x-c))(x^2 - (x-c)) = 0

Then do b^2-4ac on those

Reply 5

Lemmingz95

OP

Once you square-root both sides you're left with
plus or minus (x^2 = x - c)

given that I know there is 4 solutions, you know that each discriminant (plus OR minus) has to be greater than zero.

This gives:

1 - 4c > 0 => c <= 1/4
1 + 4c > 0 => c >= -1/4

so the answer is (b)

Reply 6

Lemmingz95

OP

Thats how I did it, but yes, it is better thankyou!

Determine where solutions exist

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