Number of integral values of $c$ in solution set
$begingroup$
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set
Try: writing quadratic equation as $$f(x)=x^2-bigg(frac{2c}{c-5}bigg)x+frac{c-4}{c-5}=0;;, cneq 5$$
$f(x)$ is upward parabola which cut $x$ axis at $(0,2)$ and other intersection in $(2,3)$
i. e $x=2$ lie between the roots means $f(2)<0$
$$f(2)=4-frac{4c}{c-5}+frac{c-4}{c-5}<0$$
$$frac{4(c-5)-4c+c-4}{c-5}<0$$
$$frac{c-24}{c-5}<0Rightarrow 5<c<24$$
I am getting integer values of $c$ are $18$
but answer given as $11$
could someone help me whats wrong in my reasoning
calculus
asked Jan 11 at 7:10
DXTDXT
5,9742732
$endgroup$
add a comment |
$begingroup$
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set
Try: writing quadratic equation as $$f(x)=x^2-bigg(frac{2c}{c-5}bigg)x+frac{c-4}{c-5}=0;;, cneq 5$$
$f(x)$ is upward parabola which cut $x$ axis at $(0,2)$ and other intersection in $(2,3)$
i. e $x=2$ lie between the roots means $f(2)<0$
$$f(2)=4-frac{4c}{c-5}+frac{c-4}{c-5}<0$$
$$frac{4(c-5)-4c+c-4}{c-5}<0$$
$$frac{c-24}{c-5}<0Rightarrow 5<c<24$$
I am getting integer values of $c$ are $18$
but answer given as $11$
could someone help me whats wrong in my reasoning
calculus
asked Jan 11 at 7:10
DXTDXT
5,9742732
$endgroup$
add a comment |
$begingroup$
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set
Try: writing quadratic equation as $$f(x)=x^2-bigg(frac{2c}{c-5}bigg)x+frac{c-4}{c-5}=0;;, cneq 5$$
$f(x)$ is upward parabola which cut $x$ axis at $(0,2)$ and other intersection in $(2,3)$
i. e $x=2$ lie between the roots means $f(2)<0$
$$f(2)=4-frac{4c}{c-5}+frac{c-4}{c-5}<0$$
$$frac{4(c-5)-4c+c-4}{c-5}<0$$
$$frac{c-24}{c-5}<0Rightarrow 5<c<24$$
I am getting integer values of $c$ are $18$
but answer given as $11$
could someone help me whats wrong in my reasoning
calculus
asked Jan 11 at 7:10
DXTDXT
5,9742732
$endgroup$
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set
Try: writing quadratic equation as $$f(x)=x^2-bigg(frac{2c}{c-5}bigg)x+frac{c-4}{c-5}=0;;, cneq 5$$
$f(x)$ is upward parabola which cut $x$ axis at $(0,2)$ and other intersection in $(2,3)$
i. e $x=2$ lie between the roots means $f(2)<0$
$$f(2)=4-frac{4c}{c-5}+frac{c-4}{c-5}<0$$
$$frac{4(c-5)-4c+c-4}{c-5}<0$$
$$frac{c-24}{c-5}<0Rightarrow 5<c<24$$
I am getting integer values of $c$ are $18$
but answer given as $11$
could someone help me whats wrong in my reasoning
calculus
calculus
asked Jan 11 at 7:10
DXTDXT
5,9742732
asked Jan 11 at 7:10
DXTDXT
5,9742732
asked Jan 11 at 7:10
DXTDXT
5,9742732
asked Jan 11 at 7:10
DXTDXT
5,9742732
asked Jan 11 at 7:10
DXTDXT
5,9742732
5,9742732
add a comment |
add a comment |
1 Answer
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$begingroup$
You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
answered Jan 11 at 7:15
ArthurArthur
117k7116200
$endgroup$
add a comment |
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$begingroup$
You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
answered Jan 11 at 7:15
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
answered Jan 11 at 7:15
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
answered Jan 11 at 7:15
ArthurArthur
117k7116200
$endgroup$
You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
answered Jan 11 at 7:15
ArthurArthur
117k7116200
edited Jan 11 at 7:21
answered Jan 11 at 7:15
ArthurArthur
117k7116200
answered Jan 11 at 7:15
ArthurArthur
117k7116200
answered Jan 11 at 7:15
ArthurArthur
117k7116200
117k7116200
add a comment |
add a comment |
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