# Homework Help: Find the value of k in the equation

1. Dec 24, 2013

### shivam01anand

1. The problem statement, all variables and given/known data

If the quadratic equations has 2 roots one of which is the square of the other find the value of k

2. Relevant equations

x^2-x+k=0

3. The attempt at a solution

let p and p^2 be the roots..

sum of roots= p(p+1)=1

product= p^3=k

then solving for p[quadratic] and then substituting into p^3=k

but the cubing part of p is rather tedious and not yielding the desired answer!

Where am i wrong
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 24, 2013

### BOAS

Are you familiar with these relationships?;

α + β = -b/a

αβ = c/a

Where α and β are the roots of the quadratic and a, b, c correspond to the coefficients (ax^2 + bx + c)

3. Dec 24, 2013

### haruspex

That's what was used in the OP.

4. Dec 24, 2013

### haruspex

There is a better way than cubing p. Use the two equations you have to eliminate the higher powers of p successively.

5. Dec 24, 2013

### BOAS

Right you are.

I missed that on account of both roots using p.

6. Dec 24, 2013

### shivam01anand

hmm okay a better way.

p^3=k

p^3-1=k-1

(p-1)(p^2+1+p)=k-1

(-P^2)(-2)=k-1

so that reduces p^3 to p^2 alright :D

have i cracked the quest for the simpler way???

7. Dec 24, 2013

### HallsofIvy

It seems to me that simplest is
$$(x- p)(x- p^2)= x^2- x+ k$$
$$x^2- (p+ p^2)x+ p^3= x^2- x+ k$$

So $k= p^3$ and $p+ p^2= 1$

Use $p+ p^2= 1$ to determine p and then $k= p^3$ to determine k.

8. Dec 24, 2013

### shivam01anand

hey did we not get that directly from sum of roots and product of roots.

the question was whether you could avoid doing the cubing part..

thanksss

9. Dec 24, 2013

### Saitama

Hi Shivam!

You have $p^3=k$. Rewrite it as $p^3-1=k-1$. Does that give you a hint? ;)

EDIT: Okay, I see you have done this.
Good so far but I don't understand what you did next. What is the value of $p^2+p+1$?

10. Dec 24, 2013

### shivam01anand

write that as -2 :D

11. Dec 24, 2013

### Saitama

It is not -2, check again!

12. Dec 24, 2013

### shivam01anand

Okay, +2 then!

13. Dec 24, 2013

### Saitama

Yes, so have you solved the problem?

14. Dec 24, 2013

### shivam01anand

yes that's why i stopped commenting
;p

15. Dec 24, 2013

### shivam01anand

yes that's why i stopped commenting
;p

16. Dec 24, 2013

### haruspex

Fwiw, what I had in mind was:
$p^2=1-p$
$p^3=p-p^2=p - (1-p)$
$k = p^3 = 2p-1$

17. Dec 25, 2013

### kastelian

so in the end there are two legitimate values of k, is that correct ?

18. Dec 25, 2013

### shivam01anand

Haa this looks just like the reducing in matrices determints. using properties.
Will surely keep this "trick" in mind

19. Dec 26, 2013

### ehild

$$x_{1,2}=\frac{1\pm\sqrt{1-4k}}{2}$$

x1=x22---> $$\frac{1+\sqrt{1-4k}}{2}=\left(\frac{1-\sqrt{1-4k}}{2}\right)^2$$
Expanding, simplifying leads to $\sqrt{1-4k}=-k$, that means k<0.

k=-2-√5.
The roots are 0.5(√5+3 )and -0.5(√5+1).

ehild