roots of cubic equation complex












1















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







share|cite|improve this question
























  • Are you looking for somewhere to start, or have you already worked out some of this problem?
    – Toby Mak
    Nov 23 at 6:03


















1















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







share|cite|improve this question
























  • Are you looking for somewhere to start, or have you already worked out some of this problem?
    – Toby Mak
    Nov 23 at 6:03
















1












1








1


1






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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • 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












1 Answer
1






active

oldest

votes


















1














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 $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$






share|cite|improve this answer





















  • 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











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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 $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$






share|cite|improve this answer





















  • 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
















1














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 $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$






share|cite|improve this answer





















  • 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














1












1








1






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 $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$






share|cite|improve this answer












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 $kin(a+sqrt{b}, b(a+color{red}{sqrt{8}}))$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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