Mixing
find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$
Clash Royale CLAN TAG#URR8PPP
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I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $sqrt(x-1)^2 + (-1 -1 + x)^2=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = pmsqrt1.75$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
algebra-precalculus
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
asked Jul 19 at 23:38
maybedave
746
 |Â
show 7 more comments
up vote
1
down vote
favorite
I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $sqrt(x-1)^2 + (-1 -1 + x)^2=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = pmsqrt1.75$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
algebra-precalculus
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
asked Jul 19 at 23:38
maybedave
746
Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line.
– amd
Jul 19 at 23:41
I did this but the points don't show up on the line. This is why I know I'm doing something wrong.
– maybedave
Jul 19 at 23:46
In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.)
– amd
Jul 19 at 23:47
yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x
– maybedave
Jul 19 at 23:48
If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$.
– David
Jul 19 at 23:50
 |Â
show 7 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $sqrt(x-1)^2 + (-1 -1 + x)^2=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = pmsqrt1.75$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
algebra-precalculus
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
asked Jul 19 at 23:38
maybedave
746
I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $sqrt(x-1)^2 + (-1 -1 + x)^2=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = pmsqrt1.75$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
algebra-precalculus
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
asked Jul 19 at 23:38
maybedave
746
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
edited Jul 20 at 8:08
N. F. Taussig
38.2k93053
38.2k93053
asked Jul 19 at 23:38
maybedave
746
asked Jul 19 at 23:38
maybedave
746
asked Jul 19 at 23:38
maybedave
746
746
Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line.
– amd
Jul 19 at 23:41
I did this but the points don't show up on the line. This is why I know I'm doing something wrong.
– maybedave
Jul 19 at 23:46
In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.)
– amd
Jul 19 at 23:47
yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x
– maybedave
Jul 19 at 23:48
If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$.
– David
Jul 19 at 23:50
 |Â
show 7 more comments
Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line.
– amd
Jul 19 at 23:41
I did this but the points don't show up on the line. This is why I know I'm doing something wrong.
– maybedave
Jul 19 at 23:46
In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.)
– amd
Jul 19 at 23:47
yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x
– maybedave
Jul 19 at 23:48
If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$.
– David
Jul 19 at 23:50
Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line.
– amd
Jul 19 at 23:41
Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line.
– amd
Jul 19 at 23:41
I did this but the points don't show up on the line. This is why I know I'm doing something wrong.
– maybedave
Jul 19 at 23:46
I did this but the points don't show up on the line. This is why I know I'm doing something wrong.
– maybedave
Jul 19 at 23:46
In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.)
– amd
Jul 19 at 23:47
In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.)
– amd
Jul 19 at 23:47
yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x
– maybedave
Jul 19 at 23:48
yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x
– maybedave
Jul 19 at 23:48
If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$.
– David
Jul 19 at 23:50
If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$.
– David
Jul 19 at 23:50
 |Â
show 7 more comments
5 Answers
5
active
oldest
votes
up vote
0
down vote
accepted
Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $left(frac32+fracsqrt72,-frac12+fracsqrt72right)$ and $left(frac32-fracsqrt72,-frac12-fracsqrt72right)$.
answered Jul 20 at 0:03
amd
25.9k2943
add a comment |Â
up vote
2
down vote
Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $frac32pmfracsqrt 72$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(frac32pmfracsqrt 72,-frac12pmfracsqrt 72)$
answered Jul 19 at 23:49
RayDansh
884214
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
add a comment |Â
up vote
2
down vote
Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=dfrac-bpmsqrtb^2-4ac2a$$
$$x=dfrac32pmdfrac12sqrt7,$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
answered Jul 19 at 23:47
Key Flex
4,336425
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
 |Â
show 5 more comments
up vote
0
down vote
You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say
$x = frac 3pmsqrt 72$ is simpler than $x = 1.5 pm sqrt 1.75$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - frac 3+sqrt 72 = frac -1-sqrt 72\y = 1 - frac 3-sqrt 72 = frac -1+sqrt 72$
answered Jul 19 at 23:55
Doug M
39.2k31749
add a comment |Â
up vote
0
down vote
After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= frac-1pm sqrt72.$$
answered Jul 20 at 8:30
Narasimham
20.2k51957
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $left(frac32+fracsqrt72,-frac12+fracsqrt72right)$ and $left(frac32-fracsqrt72,-frac12-fracsqrt72right)$.
answered Jul 20 at 0:03
amd
25.9k2943
add a comment |Â
up vote
0
down vote
accepted
Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $left(frac32+fracsqrt72,-frac12+fracsqrt72right)$ and $left(frac32-fracsqrt72,-frac12-fracsqrt72right)$.
answered Jul 20 at 0:03
amd
25.9k2943
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $left(frac32+fracsqrt72,-frac12+fracsqrt72right)$ and $left(frac32-fracsqrt72,-frac12-fracsqrt72right)$.
answered Jul 20 at 0:03
amd
25.9k2943
Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $left(frac32+fracsqrt72,-frac12+fracsqrt72right)$ and $left(frac32-fracsqrt72,-frac12-fracsqrt72right)$.
answered Jul 20 at 0:03
amd
25.9k2943
answered Jul 20 at 0:03
amd
25.9k2943
answered Jul 20 at 0:03
amd
25.9k2943
answered Jul 20 at 0:03
amd
25.9k2943
25.9k2943
add a comment |Â
add a comment |Â
up vote
2
down vote
Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $frac32pmfracsqrt 72$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(frac32pmfracsqrt 72,-frac12pmfracsqrt 72)$
answered Jul 19 at 23:49
RayDansh
884214
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
add a comment |Â
up vote
2
down vote
Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $frac32pmfracsqrt 72$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(frac32pmfracsqrt 72,-frac12pmfracsqrt 72)$
answered Jul 19 at 23:49
RayDansh
884214
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $frac32pmfracsqrt 72$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(frac32pmfracsqrt 72,-frac12pmfracsqrt 72)$
answered Jul 19 at 23:49
RayDansh
884214
Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $frac32pmfracsqrt 72$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(frac32pmfracsqrt 72,-frac12pmfracsqrt 72)$
answered Jul 19 at 23:49
RayDansh
884214
answered Jul 19 at 23:49
RayDansh
884214
answered Jul 19 at 23:49
RayDansh
884214
answered Jul 19 at 23:49
RayDansh
884214
884214
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
add a comment |Â
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation.
– amd
Jul 19 at 23:52
add a comment |Â
up vote
2
down vote
Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=dfrac-bpmsqrtb^2-4ac2a$$
$$x=dfrac32pmdfrac12sqrt7,$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
answered Jul 19 at 23:47
Key Flex
4,336425
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
 |Â
show 5 more comments
up vote
2
down vote
Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=dfrac-bpmsqrtb^2-4ac2a$$
$$x=dfrac32pmdfrac12sqrt7,$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
answered Jul 19 at 23:47
Key Flex
4,336425
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
 |Â
show 5 more comments
up vote
2
down vote
up vote
2
down vote
Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=dfrac-bpmsqrtb^2-4ac2a$$
$$x=dfrac32pmdfrac12sqrt7,$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
answered Jul 19 at 23:47
Key Flex
4,336425
Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=dfrac-bpmsqrtb^2-4ac2a$$
$$x=dfrac32pmdfrac12sqrt7,$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
answered Jul 19 at 23:47
Key Flex
4,336425
edited Jul 20 at 0:03
answered Jul 19 at 23:47
Key Flex
4,336425
answered Jul 19 at 23:47
Key Flex
4,336425
answered Jul 19 at 23:47
Key Flex
4,336425
4,336425
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
 |Â
show 5 more comments
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution.
– amd
Jul 19 at 23:49
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
The $x$ values which the O.P found were wrong.
– Key Flex
Jul 19 at 23:50
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
@Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form.
– maybedave
Jul 19 at 23:51
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
I was hoping someone could tell me what steps I did wrong though in the O.P.
– maybedave
Jul 19 at 23:52
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
The values are correct to the two decimal places used.
– amd
Jul 19 at 23:54
 |Â
show 5 more comments
up vote
0
down vote
You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say
$x = frac 3pmsqrt 72$ is simpler than $x = 1.5 pm sqrt 1.75$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - frac 3+sqrt 72 = frac -1-sqrt 72\y = 1 - frac 3-sqrt 72 = frac -1+sqrt 72$
answered Jul 19 at 23:55
Doug M
39.2k31749
add a comment |Â
up vote
0
down vote
You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say
$x = frac 3pmsqrt 72$ is simpler than $x = 1.5 pm sqrt 1.75$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - frac 3+sqrt 72 = frac -1-sqrt 72\y = 1 - frac 3-sqrt 72 = frac -1+sqrt 72$
answered Jul 19 at 23:55
Doug M
39.2k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say
$x = frac 3pmsqrt 72$ is simpler than $x = 1.5 pm sqrt 1.75$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - frac 3+sqrt 72 = frac -1-sqrt 72\y = 1 - frac 3-sqrt 72 = frac -1+sqrt 72$
answered Jul 19 at 23:55
Doug M
39.2k31749
You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say
$x = frac 3pmsqrt 72$ is simpler than $x = 1.5 pm sqrt 1.75$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - frac 3+sqrt 72 = frac -1-sqrt 72\y = 1 - frac 3-sqrt 72 = frac -1+sqrt 72$
answered Jul 19 at 23:55
Doug M
39.2k31749
edited Jul 20 at 0:00
answered Jul 19 at 23:55
Doug M
39.2k31749
answered Jul 19 at 23:55
Doug M
39.2k31749
answered Jul 19 at 23:55
Doug M
39.2k31749
39.2k31749
add a comment |Â
add a comment |Â
up vote
0
down vote
After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= frac-1pm sqrt72.$$
answered Jul 20 at 8:30
Narasimham
20.2k51957
add a comment |Â
up vote
0
down vote
After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= frac-1pm sqrt72.$$
answered Jul 20 at 8:30
Narasimham
20.2k51957
add a comment |Â
up vote
0
down vote
up vote
0
down vote
After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= frac-1pm sqrt72.$$
answered Jul 20 at 8:30
Narasimham
20.2k51957
After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= frac-1pm sqrt72.$$
answered Jul 20 at 8:30
Narasimham
20.2k51957
answered Jul 20 at 8:30
Narasimham
20.2k51957
answered Jul 20 at 8:30
Narasimham
20.2k51957
answered Jul 20 at 8:30
Narasimham
20.2k51957
20.2k51957
add a comment |Â
add a comment |Â
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Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line.
– amd
Jul 19 at 23:41
I did this but the points don't show up on the line. This is why I know I'm doing something wrong.
– maybedave
Jul 19 at 23:46
In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.)
– amd
Jul 19 at 23:47
yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x
– maybedave
Jul 19 at 23:48
If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$.
– David
Jul 19 at 23:50