Minimum possible number of positive root of the quadratic equation $x^2-(1+lambda)x+lambda-2=0,lambdain R$ is
Minimum possible number of positive root of the quadratic equation $x^2-(1+lambda)x+lambda-2=0,lambdain R$ is
$(a)2$
$(b)1$
$(c)0$
$(d)3$
$x^2-(1+lambda)x+lambda-2=0$ I changed this equation to $lambda=frac{x^2-x-2}{x-1}$.I am stuck now.
quadratics
asked Dec 28 '18 at 12:40
user984325user984325
246112
add a comment |
Minimum possible number of positive root of the quadratic equation $x^2-(1+lambda)x+lambda-2=0,lambdain R$ is
$(a)2$
$(b)1$
$(c)0$
$(d)3$
$x^2-(1+lambda)x+lambda-2=0$ I changed this equation to $lambda=frac{x^2-x-2}{x-1}$.I am stuck now.
quadratics
asked Dec 28 '18 at 12:40
user984325user984325
246112
What do you mean by "Minimum possible number of possible root of the quadratic equation"? The minimum number of values $lambda$ can take on? The minimum value it can take on? And what is $R$ here? The real numbers? There's too much content not present.
– Eevee Trainer
Dec 28 '18 at 12:42
R is real numbers.
– user984325
Dec 28 '18 at 13:06
add a comment |
Minimum possible number of positive root of the quadratic equation $x^2-(1+lambda)x+lambda-2=0,lambdain R$ is
$(a)2$
$(b)1$
$(c)0$
$(d)3$
$x^2-(1+lambda)x+lambda-2=0$ I changed this equation to $lambda=frac{x^2-x-2}{x-1}$.I am stuck now.
quadratics
asked Dec 28 '18 at 12:40
user984325user984325
246112
Minimum possible number of positive root of the quadratic equation $x^2-(1+lambda)x+lambda-2=0,lambdain R$ is
$(a)2$
$(b)1$
$(c)0$
$(d)3$
$x^2-(1+lambda)x+lambda-2=0$ I changed this equation to $lambda=frac{x^2-x-2}{x-1}$.I am stuck now.
quadratics
quadratics
asked Dec 28 '18 at 12:40
user984325user984325
246112
asked Dec 28 '18 at 12:40
user984325user984325
246112
edited Dec 28 '18 at 13:30
user984325
asked Dec 28 '18 at 12:40
user984325user984325
246112
asked Dec 28 '18 at 12:40
user984325user984325
246112
asked Dec 28 '18 at 12:40
user984325user984325
246112
246112
What do you mean by "Minimum possible number of possible root of the quadratic equation"? The minimum number of values $lambda$ can take on? The minimum value it can take on? And what is $R$ here? The real numbers? There's too much content not present.
– Eevee Trainer
Dec 28 '18 at 12:42
R is real numbers.
– user984325
Dec 28 '18 at 13:06
add a comment |
What do you mean by "Minimum possible number of possible root of the quadratic equation"? The minimum number of values $lambda$ can take on? The minimum value it can take on? And what is $R$ here? The real numbers? There's too much content not present.
– Eevee Trainer
Dec 28 '18 at 12:42
R is real numbers.
– user984325
Dec 28 '18 at 13:06
What do you mean by "Minimum possible number of possible root of the quadratic equation"? The minimum number of values $lambda$ can take on? The minimum value it can take on? And what is $R$ here? The real numbers? There's too much content not present.
– Eevee Trainer
Dec 28 '18 at 12:42
What do you mean by "Minimum possible number of possible root of the quadratic equation"? The minimum number of values $lambda$ can take on? The minimum value it can take on? And what is $R$ here? The real numbers? There's too much content not present.
– Eevee Trainer
Dec 28 '18 at 12:42
R is real numbers.
– user984325
Dec 28 '18 at 13:06
R is real numbers.
– user984325
Dec 28 '18 at 13:06
add a comment |
3 Answers
3
active
oldest
votes
The discriminant is positive, so the equation has distinct roots for every $lambda$.
The arithmetic mean of the roots is $(lambda+1)/2$. Thus $lambdale-1$ implies at least a root is negative.
Can both roots be negative? Hint: Descartes' rule of signs.
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
add a comment |
The solutions are
$$x=frac{lambda+1pm sqrt{(lambda+1)^2-4(lambda-2)}}{2}.$$
The factor of 2 plays no role so we can just focus on the numerator. The discriminant is
$$(lambda+1)^2-4(lambda-2) = lambda^2-2lambda+9=(lambda-1)^2+8.$$
This is positive, so there are two real roots. Working with the root coming from +discriminant,
$$lambda+1+sqrt{(lambda-1)^2+8} ge lambda+1+|lambda-1|. $$
This is never negative since if $lambda+1$ is negative, $lambda-1$ is more negative and thus its absolute value is more positive. Therefore there is at least one positive root.
The -discriminant case is more fun.
$$lambda+1-sqrt{(lambda-1)^2+8} le lambda+1-sqrt{8}.$$
Since the right half can be negative, we see that for at least some choices of $lambda$ ($lambdale sqrt{8}-1$), this is negative, giving that the minimum number of positive roots is one.
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
add a comment |
If $lambda = 2 $ then $ x = 0 $ is a non negative solution.
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The discriminant is positive, so the equation has distinct roots for every $lambda$.
The arithmetic mean of the roots is $(lambda+1)/2$. Thus $lambdale-1$ implies at least a root is negative.
Can both roots be negative? Hint: Descartes' rule of signs.
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
add a comment |
The discriminant is positive, so the equation has distinct roots for every $lambda$.
The arithmetic mean of the roots is $(lambda+1)/2$. Thus $lambdale-1$ implies at least a root is negative.
Can both roots be negative? Hint: Descartes' rule of signs.
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
add a comment |
The discriminant is positive, so the equation has distinct roots for every $lambda$.
The arithmetic mean of the roots is $(lambda+1)/2$. Thus $lambdale-1$ implies at least a root is negative.
Can both roots be negative? Hint: Descartes' rule of signs.
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
The discriminant is positive, so the equation has distinct roots for every $lambda$.
The arithmetic mean of the roots is $(lambda+1)/2$. Thus $lambdale-1$ implies at least a root is negative.
Can both roots be negative? Hint: Descartes' rule of signs.
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
answered Dec 28 '18 at 14:16
egregegreg
179k1484202
179k1484202
add a comment |
add a comment |
The solutions are
$$x=frac{lambda+1pm sqrt{(lambda+1)^2-4(lambda-2)}}{2}.$$
The factor of 2 plays no role so we can just focus on the numerator. The discriminant is
$$(lambda+1)^2-4(lambda-2) = lambda^2-2lambda+9=(lambda-1)^2+8.$$
This is positive, so there are two real roots. Working with the root coming from +discriminant,
$$lambda+1+sqrt{(lambda-1)^2+8} ge lambda+1+|lambda-1|. $$
This is never negative since if $lambda+1$ is negative, $lambda-1$ is more negative and thus its absolute value is more positive. Therefore there is at least one positive root.
The -discriminant case is more fun.
$$lambda+1-sqrt{(lambda-1)^2+8} le lambda+1-sqrt{8}.$$
Since the right half can be negative, we see that for at least some choices of $lambda$ ($lambdale sqrt{8}-1$), this is negative, giving that the minimum number of positive roots is one.
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
add a comment |
The solutions are
$$x=frac{lambda+1pm sqrt{(lambda+1)^2-4(lambda-2)}}{2}.$$
The factor of 2 plays no role so we can just focus on the numerator. The discriminant is
$$(lambda+1)^2-4(lambda-2) = lambda^2-2lambda+9=(lambda-1)^2+8.$$
This is positive, so there are two real roots. Working with the root coming from +discriminant,
$$lambda+1+sqrt{(lambda-1)^2+8} ge lambda+1+|lambda-1|. $$
This is never negative since if $lambda+1$ is negative, $lambda-1$ is more negative and thus its absolute value is more positive. Therefore there is at least one positive root.
The -discriminant case is more fun.
$$lambda+1-sqrt{(lambda-1)^2+8} le lambda+1-sqrt{8}.$$
Since the right half can be negative, we see that for at least some choices of $lambda$ ($lambdale sqrt{8}-1$), this is negative, giving that the minimum number of positive roots is one.
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
add a comment |
The solutions are
$$x=frac{lambda+1pm sqrt{(lambda+1)^2-4(lambda-2)}}{2}.$$
The factor of 2 plays no role so we can just focus on the numerator. The discriminant is
$$(lambda+1)^2-4(lambda-2) = lambda^2-2lambda+9=(lambda-1)^2+8.$$
This is positive, so there are two real roots. Working with the root coming from +discriminant,
$$lambda+1+sqrt{(lambda-1)^2+8} ge lambda+1+|lambda-1|. $$
This is never negative since if $lambda+1$ is negative, $lambda-1$ is more negative and thus its absolute value is more positive. Therefore there is at least one positive root.
The -discriminant case is more fun.
$$lambda+1-sqrt{(lambda-1)^2+8} le lambda+1-sqrt{8}.$$
Since the right half can be negative, we see that for at least some choices of $lambda$ ($lambdale sqrt{8}-1$), this is negative, giving that the minimum number of positive roots is one.
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
The solutions are
$$x=frac{lambda+1pm sqrt{(lambda+1)^2-4(lambda-2)}}{2}.$$
The factor of 2 plays no role so we can just focus on the numerator. The discriminant is
$$(lambda+1)^2-4(lambda-2) = lambda^2-2lambda+9=(lambda-1)^2+8.$$
This is positive, so there are two real roots. Working with the root coming from +discriminant,
$$lambda+1+sqrt{(lambda-1)^2+8} ge lambda+1+|lambda-1|. $$
This is never negative since if $lambda+1$ is negative, $lambda-1$ is more negative and thus its absolute value is more positive. Therefore there is at least one positive root.
The -discriminant case is more fun.
$$lambda+1-sqrt{(lambda-1)^2+8} le lambda+1-sqrt{8}.$$
Since the right half can be negative, we see that for at least some choices of $lambda$ ($lambdale sqrt{8}-1$), this is negative, giving that the minimum number of positive roots is one.
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
edited Dec 28 '18 at 14:37
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
answered Dec 28 '18 at 14:03
Cameron WilliamsCameron Williams
22.3k43579
22.3k43579
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
add a comment |
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
You could do the -
– Cameron Williams
Dec 28 '18 at 14:53
add a comment |
If $lambda = 2 $ then $ x = 0 $ is a non negative solution.
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
add a comment |
If $lambda = 2 $ then $ x = 0 $ is a non negative solution.
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
add a comment |
If $lambda = 2 $ then $ x = 0 $ is a non negative solution.
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
If $lambda = 2 $ then $ x = 0 $ is a non negative solution.
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
answered Dec 28 '18 at 13:57
Angel MorenoAngel Moreno
37915
37915
add a comment |
add a comment |
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What do you mean by "Minimum possible number of possible root of the quadratic equation"? The minimum number of values $lambda$ can take on? The minimum value it can take on? And what is $R$ here? The real numbers? There's too much content not present.
– Eevee Trainer
Dec 28 '18 at 12:42
R is real numbers.
– user984325
Dec 28 '18 at 13:06