Find the value of k in the equation

[shivam01anand]

Homework Statement

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

Homework Equations

x^2-x+k=0

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

 

[BOAS]

Homework Statement

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

Homework Equations

x^2-x+k=0

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

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)

 

[haruspex]

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)

That's what was used in the OP.

 

[haruspex]

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

Please post your answer and the given answer.
There is a better way than cubing p. Use the two equations you have to eliminate the higher powers of p successively.

 

[BOAS]

That's what was used in the OP.

Right you are.

I missed that on account of both roots using p.

 

[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?

 

[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.

 

[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

 

[Saitama]

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

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.

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

Good so far but I don't understand what you did next. What is the value of $p^2+p+1$?

 

[shivam01anand]

Hi Shivam!



Good so far but I don't understand what you did next. What is the value of $p^2+p+1$?

write that as -2 :D

 

[Saitama]

write that as -2 :D

It is not -2, check again!

 

[shivam01anand]

Okay, +2 then!

 

[Saitama]

Okay, +2 then!

Yes, so have you solved the problem?

 

[shivam01anand]

yes that's why i stopped commenting
;p

 

[shivam01anand]

yes that's why i stopped commenting
;p

 

[haruspex]

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?

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

 

[kastelian]

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

 

[shivam01anand]

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

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

 

[ehild]

You can also start with the quadratic formula.

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

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.

Spoiler

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

ehild