Condition on $lambda$ so that the equation is satisfied
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Let there exist real numbers $a,b,c,d$ such that $lambda x^2+2xy+y^2=(ax+by)^2+(cx+dy)^2 forall x,yinmathbb{R}$, then what is the condition on $lambda$?
I think such a $lambda$ does not exist. I got the equations $$begin{aligned} ab+cd&=1\ a^2+c^2&=lambda \ b^2+d^2&=1end{aligned}$$ but am unable to proceed further. Any hints? Thanks beforehand.
algebra-precalculus polynomials
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
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add a comment |
$begingroup$
Let there exist real numbers $a,b,c,d$ such that $lambda x^2+2xy+y^2=(ax+by)^2+(cx+dy)^2 forall x,yinmathbb{R}$, then what is the condition on $lambda$?
I think such a $lambda$ does not exist. I got the equations $$begin{aligned} ab+cd&=1\ a^2+c^2&=lambda \ b^2+d^2&=1end{aligned}$$ but am unable to proceed further. Any hints? Thanks beforehand.
algebra-precalculus polynomials
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
$endgroup$
add a comment |
$begingroup$
Let there exist real numbers $a,b,c,d$ such that $lambda x^2+2xy+y^2=(ax+by)^2+(cx+dy)^2 forall x,yinmathbb{R}$, then what is the condition on $lambda$?
I think such a $lambda$ does not exist. I got the equations $$begin{aligned} ab+cd&=1\ a^2+c^2&=lambda \ b^2+d^2&=1end{aligned}$$ but am unable to proceed further. Any hints? Thanks beforehand.
algebra-precalculus polynomials
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
$endgroup$
Let there exist real numbers $a,b,c,d$ such that $lambda x^2+2xy+y^2=(ax+by)^2+(cx+dy)^2 forall x,yinmathbb{R}$, then what is the condition on $lambda$?
I think such a $lambda$ does not exist. I got the equations $$begin{aligned} ab+cd&=1\ a^2+c^2&=lambda \ b^2+d^2&=1end{aligned}$$ but am unable to proceed further. Any hints? Thanks beforehand.
algebra-precalculus polynomials
algebra-precalculus polynomials
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
edited Dec 12 '18 at 11:14
vidyarthi
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
asked Dec 12 '18 at 10:55
vidyarthividyarthi
2,9641832
2,9641832
add a comment |
add a comment |
1 Answer
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Hint the right-hand side is never negative
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Hint the right-hand side is never negative
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
add a comment |
$begingroup$
Hint the right-hand side is never negative
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
add a comment |
$begingroup$
Hint the right-hand side is never negative
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
$endgroup$
Hint the right-hand side is never negative
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
answered Dec 12 '18 at 11:03
Empy2Empy2
33.5k12261
33.5k12261
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
add a comment |
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
I think it should be $ax+by$ and $cx+dy$
$endgroup$
– Empy2
Dec 12 '18 at 11:11
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
$begingroup$
yes, thanks, edited.
$endgroup$
– vidyarthi
Dec 12 '18 at 11:15
add a comment |
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