Finding x, given the terms of an arithmetic sequence
So the question is this:
Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:
begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}
I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:
Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.
begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.
Working out the values of $U_2-U_1$ and $U_3-U_2$:
begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}
That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?
sequences-and-series arithmetic
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
asked Jun 3 '17 at 18:12
Juan Diego Last
12
add a comment |
So the question is this:
Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:
begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}
I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:
Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.
begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.
Working out the values of $U_2-U_1$ and $U_3-U_2$:
begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}
That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?
sequences-and-series arithmetic
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
asked Jun 3 '17 at 18:12
Juan Diego Last
12
Seems good to me.
– tjeremie
Jun 3 '17 at 18:17
The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22
You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28
Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16
@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23
add a comment |
So the question is this:
Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:
begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}
I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:
Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.
begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.
Working out the values of $U_2-U_1$ and $U_3-U_2$:
begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}
That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?
sequences-and-series arithmetic
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
asked Jun 3 '17 at 18:12
Juan Diego Last
12
So the question is this:
Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:
begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}
I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:
Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.
begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.
Working out the values of $U_2-U_1$ and $U_3-U_2$:
begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}
That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?
sequences-and-series arithmetic
sequences-and-series arithmetic
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
asked Jun 3 '17 at 18:12
Juan Diego Last
12
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
asked Jun 3 '17 at 18:12
Juan Diego Last
12
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
edited Jun 3 '17 at 18:24
John Wayland Bales
13.8k21137
13.8k21137
asked Jun 3 '17 at 18:12
Juan Diego Last
12
asked Jun 3 '17 at 18:12
Juan Diego Last
12
asked Jun 3 '17 at 18:12
Juan Diego Last
12
12
Seems good to me.
– tjeremie
Jun 3 '17 at 18:17
The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22
You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28
Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16
@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23
add a comment |
Seems good to me.
– tjeremie
Jun 3 '17 at 18:17
The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22
You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28
Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16
@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23
Seems good to me.
– tjeremie
Jun 3 '17 at 18:17
Seems good to me.
– tjeremie
Jun 3 '17 at 18:17
The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22
The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22
You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28
You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28
Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16
Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16
@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23
@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23
add a comment |
2 Answers
2
active
oldest
votes
Let's see:
$2x^4−68x^2−1468=x^4−8x^2−2007$
$y = x^2$
$2y^2-68y-1468=y^2-8y-2007$
$y^2-60y-539=0$
$(y-30)^2-539=900$
$(y-30)^2=1439$
$y-30=pm 1439^{1/2}$
$y = 30pm 1439^{1/2}$
$x = 1469^{1/2}$
$x = 1409^{1/2}$
Pressed enter accidentally up there and it looks weird as a comment.
edited Jun 3 '17 at 19:45
callculus
17.8k31427
answered Jun 3 '17 at 19:25
Juan Diego Last
12
add a comment |
Since $ 2007-1468=539$ you have
$(y-30)^2color{red}+539=900quad |-539$
$(y-30)^2=361quad | sqrt{()} $
$y-30=pm 19$
$
y_1=11, y_2=49$
Re-subtitution $sqrt{y}=pm x$
$x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$
answered Jun 3 '17 at 19:59
callculus
17.8k31427
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let's see:
$2x^4−68x^2−1468=x^4−8x^2−2007$
$y = x^2$
$2y^2-68y-1468=y^2-8y-2007$
$y^2-60y-539=0$
$(y-30)^2-539=900$
$(y-30)^2=1439$
$y-30=pm 1439^{1/2}$
$y = 30pm 1439^{1/2}$
$x = 1469^{1/2}$
$x = 1409^{1/2}$
Pressed enter accidentally up there and it looks weird as a comment.
edited Jun 3 '17 at 19:45
callculus
17.8k31427
answered Jun 3 '17 at 19:25
Juan Diego Last
12
add a comment |
Let's see:
$2x^4−68x^2−1468=x^4−8x^2−2007$
$y = x^2$
$2y^2-68y-1468=y^2-8y-2007$
$y^2-60y-539=0$
$(y-30)^2-539=900$
$(y-30)^2=1439$
$y-30=pm 1439^{1/2}$
$y = 30pm 1439^{1/2}$
$x = 1469^{1/2}$
$x = 1409^{1/2}$
Pressed enter accidentally up there and it looks weird as a comment.
edited Jun 3 '17 at 19:45
callculus
17.8k31427
answered Jun 3 '17 at 19:25
Juan Diego Last
12
add a comment |
Let's see:
$2x^4−68x^2−1468=x^4−8x^2−2007$
$y = x^2$
$2y^2-68y-1468=y^2-8y-2007$
$y^2-60y-539=0$
$(y-30)^2-539=900$
$(y-30)^2=1439$
$y-30=pm 1439^{1/2}$
$y = 30pm 1439^{1/2}$
$x = 1469^{1/2}$
$x = 1409^{1/2}$
Pressed enter accidentally up there and it looks weird as a comment.
edited Jun 3 '17 at 19:45
callculus
17.8k31427
answered Jun 3 '17 at 19:25
Juan Diego Last
12
Let's see:
$2x^4−68x^2−1468=x^4−8x^2−2007$
$y = x^2$
$2y^2-68y-1468=y^2-8y-2007$
$y^2-60y-539=0$
$(y-30)^2-539=900$
$(y-30)^2=1439$
$y-30=pm 1439^{1/2}$
$y = 30pm 1439^{1/2}$
$x = 1469^{1/2}$
$x = 1409^{1/2}$
Pressed enter accidentally up there and it looks weird as a comment.
edited Jun 3 '17 at 19:45
callculus
17.8k31427
answered Jun 3 '17 at 19:25
Juan Diego Last
12
edited Jun 3 '17 at 19:45
callculus
17.8k31427
edited Jun 3 '17 at 19:45
callculus
17.8k31427
edited Jun 3 '17 at 19:45
callculus
17.8k31427
17.8k31427
answered Jun 3 '17 at 19:25
Juan Diego Last
12
answered Jun 3 '17 at 19:25
Juan Diego Last
12
answered Jun 3 '17 at 19:25
Juan Diego Last
12
12
add a comment |
add a comment |
Since $ 2007-1468=539$ you have
$(y-30)^2color{red}+539=900quad |-539$
$(y-30)^2=361quad | sqrt{()} $
$y-30=pm 19$
$
y_1=11, y_2=49$
Re-subtitution $sqrt{y}=pm x$
$x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$
answered Jun 3 '17 at 19:59
callculus
17.8k31427
add a comment |
Since $ 2007-1468=539$ you have
$(y-30)^2color{red}+539=900quad |-539$
$(y-30)^2=361quad | sqrt{()} $
$y-30=pm 19$
$
y_1=11, y_2=49$
Re-subtitution $sqrt{y}=pm x$
$x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$
answered Jun 3 '17 at 19:59
callculus
17.8k31427
add a comment |
Since $ 2007-1468=539$ you have
$(y-30)^2color{red}+539=900quad |-539$
$(y-30)^2=361quad | sqrt{()} $
$y-30=pm 19$
$
y_1=11, y_2=49$
Re-subtitution $sqrt{y}=pm x$
$x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$
answered Jun 3 '17 at 19:59
callculus
17.8k31427
Since $ 2007-1468=539$ you have
$(y-30)^2color{red}+539=900quad |-539$
$(y-30)^2=361quad | sqrt{()} $
$y-30=pm 19$
$
y_1=11, y_2=49$
Re-subtitution $sqrt{y}=pm x$
$x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$
answered Jun 3 '17 at 19:59
callculus
17.8k31427
answered Jun 3 '17 at 19:59
callculus
17.8k31427
answered Jun 3 '17 at 19:59
callculus
17.8k31427
answered Jun 3 '17 at 19:59
callculus
17.8k31427
17.8k31427
add a comment |
add a comment |
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Seems good to me.
– tjeremie
Jun 3 '17 at 18:17
The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22
You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28
Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16
@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23