Value of a fraction independent of x

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For $a=4$ it is known that the value of the fraction $frac(a+2)x + a^2 - 1ax-2a + 18$ is independent of $x$. The other value of $a$ for which this is the case, belong to the interval _______.

My approach : Since the fraction is independent of x for $a=4$, so we can assume that the value becomes $frac1510$ = $frac32$, by substituting $x = 0$. I don't know how to approach to the next step after this. Please help.

linear-algebra fractions

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asked Jul 27 at 8:39

MathsLearner

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For $a=4$ it is known that the value of the fraction $frac(a+2)x + a^2 - 1ax-2a + 18$ is independent of $x$. The other value of $a$ for which this is the case, belong to the interval _______.

My approach : Since the fraction is independent of x for $a=4$, so we can assume that the value becomes $frac1510$ = $frac32$, by substituting $x = 0$. I don't know how to approach to the next step after this. Please help.

linear-algebra fractions

share|cite|improve this question

asked Jul 27 at 8:39

MathsLearner

657213

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

For $a=4$ it is known that the value of the fraction $frac(a+2)x + a^2 - 1ax-2a + 18$ is independent of $x$. The other value of $a$ for which this is the case, belong to the interval _______.

My approach : Since the fraction is independent of x for $a=4$, so we can assume that the value becomes $frac1510$ = $frac32$, by substituting $x = 0$. I don't know how to approach to the next step after this. Please help.

linear-algebra fractions

share|cite|improve this question

asked Jul 27 at 8:39

MathsLearner

657213

For $a=4$ it is known that the value of the fraction $frac(a+2)x + a^2 - 1ax-2a + 18$ is independent of $x$. The other value of $a$ for which this is the case, belong to the interval _______.

My approach : Since the fraction is independent of x for $a=4$, so we can assume that the value becomes $frac1510$ = $frac32$, by substituting $x = 0$. I don't know how to approach to the next step after this. Please help.

linear-algebra fractions

share|cite|improve this question

asked Jul 27 at 8:39

MathsLearner

657213

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 8:39

MathsLearner

657213

asked Jul 27 at 8:39

MathsLearner

657213

asked Jul 27 at 8:39

MathsLearner

657213

657213

add a comment | 

add a comment | 

2 Answers
2

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up vote
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accepted

If $a=4$, then your fractions is just $frac32$. Is there another $a$ for which the fractions is constant? That would mean that $fraca+2a=fraca^2-1-2a+18$, which is equivalent to $a^3+2a^2-15-36=0$. This is a cubic equation, but you already know that it has $4$ as a root. The only other root is $-3$.

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answered Jul 27 at 8:47

José Carlos Santos

113k1696173

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up vote
2
down vote

Hint

Consider the function $$f(x)=frac(a+2)x + a^2 - 1a(x-2) + 18$$ Compute its derivative to get
$$f'(x)=-frac(a-4) (a+3)^2(a (x-2)+18)^2$$

IU am sure that you can take it from here.

share|cite|improve this answer

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

If $a=4$, then your fractions is just $frac32$. Is there another $a$ for which the fractions is constant? That would mean that $fraca+2a=fraca^2-1-2a+18$, which is equivalent to $a^3+2a^2-15-36=0$. This is a cubic equation, but you already know that it has $4$ as a root. The only other root is $-3$.

share|cite|improve this answer

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

add a comment | 

up vote
2
down vote



accepted

If $a=4$, then your fractions is just $frac32$. Is there another $a$ for which the fractions is constant? That would mean that $fraca+2a=fraca^2-1-2a+18$, which is equivalent to $a^3+2a^2-15-36=0$. This is a cubic equation, but you already know that it has $4$ as a root. The only other root is $-3$.

share|cite|improve this answer

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

If $a=4$, then your fractions is just $frac32$. Is there another $a$ for which the fractions is constant? That would mean that $fraca+2a=fraca^2-1-2a+18$, which is equivalent to $a^3+2a^2-15-36=0$. This is a cubic equation, but you already know that it has $4$ as a root. The only other root is $-3$.

share|cite|improve this answer

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

If $a=4$, then your fractions is just $frac32$. Is there another $a$ for which the fractions is constant? That would mean that $fraca+2a=fraca^2-1-2a+18$, which is equivalent to $a^3+2a^2-15-36=0$. This is a cubic equation, but you already know that it has $4$ as a root. The only other root is $-3$.

share|cite|improve this answer

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

answered Jul 27 at 8:47

José Carlos Santos

113k1696173

113k1696173

add a comment | 

add a comment | 

up vote
2
down vote

Hint

Consider the function $$f(x)=frac(a+2)x + a^2 - 1a(x-2) + 18$$ Compute its derivative to get
$$f'(x)=-frac(a-4) (a+3)^2(a (x-2)+18)^2$$

IU am sure that you can take it from here.

share|cite|improve this answer

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

add a comment | 

up vote
2
down vote

Hint

Consider the function $$f(x)=frac(a+2)x + a^2 - 1a(x-2) + 18$$ Compute its derivative to get
$$f'(x)=-frac(a-4) (a+3)^2(a (x-2)+18)^2$$

IU am sure that you can take it from here.

share|cite|improve this answer

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint

Consider the function $$f(x)=frac(a+2)x + a^2 - 1a(x-2) + 18$$ Compute its derivative to get
$$f'(x)=-frac(a-4) (a+3)^2(a (x-2)+18)^2$$

IU am sure that you can take it from here.

share|cite|improve this answer

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

Hint

Consider the function $$f(x)=frac(a+2)x + a^2 - 1a(x-2) + 18$$ Compute its derivative to get
$$f'(x)=-frac(a-4) (a+3)^2(a (x-2)+18)^2$$

IU am sure that you can take it from here.

share|cite|improve this answer

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

answered Jul 27 at 8:47

Claude Leibovici

111k1055126

111k1055126

add a comment | 

add a comment | 

 

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