roots of cubic equation complex
Given the cubic equation: $$x^3-2kx^2-4kx+k^2=0.$$
If one root of the equation is less than $1$, another root is in the interval $(1,4)$ and the third root is greater than $4$, then the value of $k$ lies in the interval $left(a+sqrt{b},b(a+sqrt{6})right)$ where $a,bin mathbb{N}$.
Find the value of $(a+b)^3+(ab+2)^2$.
My approach
Let the roots be $p,q,r$
So
$$p<1$$
$$1<q<4$$
$$r>4$$
$$p+q+r = 2k$$
$$pq+qr+rp = -4k$$
$$pqr = -k^2$$
polynomials roots
add a comment |
Given the cubic equation: $$x^3-2kx^2-4kx+k^2=0.$$
If one root of the equation is less than $1$, another root is in the interval $(1,4)$ and the third root is greater than $4$, then the value of $k$ lies in the interval $left(a+sqrt{b},b(a+sqrt{6})right)$ where $a,bin mathbb{N}$.
Find the value of $(a+b)^3+(ab+2)^2$.
My approach
Let the roots be $p,q,r$
So
$$p<1$$
$$1<q<4$$
$$r>4$$
$$p+q+r = 2k$$
$$pq+qr+rp = -4k$$
$$pqr = -k^2$$
polynomials roots
Are you looking for somewhere to start, or have you already worked out some of this problem?
– Toby Mak
Nov 23 at 6:03
add a comment |
Given the cubic equation: $$x^3-2kx^2-4kx+k^2=0.$$
If one root of the equation is less than $1$, another root is in the interval $(1,4)$ and the third root is greater than $4$, then the value of $k$ lies in the interval $left(a+sqrt{b},b(a+sqrt{6})right)$ where $a,bin mathbb{N}$.
Find the value of $(a+b)^3+(ab+2)^2$.
My approach
Let the roots be $p,q,r$
So
$$p<1$$
$$1<q<4$$
$$r>4$$
$$p+q+r = 2k$$
$$pq+qr+rp = -4k$$
$$pqr = -k^2$$
polynomials roots
Given the cubic equation: $$x^3-2kx^2-4kx+k^2=0.$$
If one root of the equation is less than $1$, another root is in the interval $(1,4)$ and the third root is greater than $4$, then the value of $k$ lies in the interval $left(a+sqrt{b},b(a+sqrt{6})right)$ where $a,bin mathbb{N}$.
Find the value of $(a+b)^3+(ab+2)^2$.
My approach
Let the roots be $p,q,r$
So
$$p<1$$
$$1<q<4$$
$$r>4$$
$$p+q+r = 2k$$
$$pq+qr+rp = -4k$$
$$pqr = -k^2$$
polynomials roots
polynomials roots
edited Nov 23 at 7:34
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Nov 23 at 5:48
Argus
1165
1165
Are you looking for somewhere to start, or have you already worked out some of this problem?
– Toby Mak
Nov 23 at 6:03
add a comment |
Are you looking for somewhere to start, or have you already worked out some of this problem?
– Toby Mak
Nov 23 at 6:03
Are you looking for somewhere to start, or have you already worked out some of this problem?
– Toby Mak
Nov 23 at 6:03
Are you looking for somewhere to start, or have you already worked out some of this problem?
– Toby Mak
Nov 23 at 6:03
add a comment |
1 Answer
1
active
oldest
votes
Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f to infty$ as $x to infty$ and $f to -infty$ as $x to -infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
begin{align*}
k^2-6k+1 & > 0\
k^2-48k+64 & < 0
end{align*}
The first inequality gives $k in (-infty, 3-2sqrt{2}) cup (3+2sqrt{2}, infty)$. The second inequality gives $k in (24-16sqrt{2}, ,, 24+16sqrt{2})$. The intersection of these intervals gives
$$k in (3+2sqrt{2}, ,, 24+16sqrt{2})=(3+sqrt{8}, ,, 8(3+sqrt{8})).$$
Thus $a=3,b=8$.
Note: I think there is a possible typo in your question. It should be $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f to infty$ as $x to infty$ and $f to -infty$ as $x to -infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
begin{align*}
k^2-6k+1 & > 0\
k^2-48k+64 & < 0
end{align*}
The first inequality gives $k in (-infty, 3-2sqrt{2}) cup (3+2sqrt{2}, infty)$. The second inequality gives $k in (24-16sqrt{2}, ,, 24+16sqrt{2})$. The intersection of these intervals gives
$$k in (3+2sqrt{2}, ,, 24+16sqrt{2})=(3+sqrt{8}, ,, 8(3+sqrt{8})).$$
Thus $a=3,b=8$.
Note: I think there is a possible typo in your question. It should be $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
add a comment |
Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f to infty$ as $x to infty$ and $f to -infty$ as $x to -infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
begin{align*}
k^2-6k+1 & > 0\
k^2-48k+64 & < 0
end{align*}
The first inequality gives $k in (-infty, 3-2sqrt{2}) cup (3+2sqrt{2}, infty)$. The second inequality gives $k in (24-16sqrt{2}, ,, 24+16sqrt{2})$. The intersection of these intervals gives
$$k in (3+2sqrt{2}, ,, 24+16sqrt{2})=(3+sqrt{8}, ,, 8(3+sqrt{8})).$$
Thus $a=3,b=8$.
Note: I think there is a possible typo in your question. It should be $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
add a comment |
Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f to infty$ as $x to infty$ and $f to -infty$ as $x to -infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
begin{align*}
k^2-6k+1 & > 0\
k^2-48k+64 & < 0
end{align*}
The first inequality gives $k in (-infty, 3-2sqrt{2}) cup (3+2sqrt{2}, infty)$. The second inequality gives $k in (24-16sqrt{2}, ,, 24+16sqrt{2})$. The intersection of these intervals gives
$$k in (3+2sqrt{2}, ,, 24+16sqrt{2})=(3+sqrt{8}, ,, 8(3+sqrt{8})).$$
Thus $a=3,b=8$.
Note: I think there is a possible typo in your question. It should be $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$
Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f to infty$ as $x to infty$ and $f to -infty$ as $x to -infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
begin{align*}
k^2-6k+1 & > 0\
k^2-48k+64 & < 0
end{align*}
The first inequality gives $k in (-infty, 3-2sqrt{2}) cup (3+2sqrt{2}, infty)$. The second inequality gives $k in (24-16sqrt{2}, ,, 24+16sqrt{2})$. The intersection of these intervals gives
$$k in (3+2sqrt{2}, ,, 24+16sqrt{2})=(3+sqrt{8}, ,, 8(3+sqrt{8})).$$
Thus $a=3,b=8$.
Note: I think there is a possible typo in your question. It should be $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$
answered Nov 23 at 6:49
Anurag A
25.5k12249
25.5k12249
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
add a comment |
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
hi anurag thanx are u from iit?
– Argus
Nov 23 at 7:18
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
@Argus Long time ago :-)
– Anurag A
Nov 23 at 7:20
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
i think you are my fb friend ur full name anurag anand u work at facebook?
– Argus
Nov 23 at 7:28
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
@Argus I'm not the one who is you FB friend and I don't work at Facebook. Sorry!!
– Anurag A
Nov 23 at 7:29
add a comment |
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Are you looking for somewhere to start, or have you already worked out some of this problem?
– Toby Mak
Nov 23 at 6:03