help solve simultaneous equation
$begingroup$
I need to solve the system
$$6x+y=3tag 1 $$
$$x^2+y^2=16tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This i need to solve by completing the square.
systems-of-equations completing-the-square
edited Jan 12 '17 at 23:51
Namaste
1
asked Jan 12 '17 at 23:45
user406484user406484
111
$endgroup$
add a comment |
$begingroup$
I need to solve the system
$$6x+y=3tag 1 $$
$$x^2+y^2=16tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This i need to solve by completing the square.
systems-of-equations completing-the-square
edited Jan 12 '17 at 23:51
Namaste
1
asked Jan 12 '17 at 23:45
user406484user406484
111
$endgroup$
$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46
$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50
add a comment |
$begingroup$
I need to solve the system
$$6x+y=3tag 1 $$
$$x^2+y^2=16tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This i need to solve by completing the square.
systems-of-equations completing-the-square
edited Jan 12 '17 at 23:51
Namaste
1
asked Jan 12 '17 at 23:45
user406484user406484
111
$endgroup$
I need to solve the system
$$6x+y=3tag 1 $$
$$x^2+y^2=16tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This i need to solve by completing the square.
systems-of-equations completing-the-square
systems-of-equations completing-the-square
edited Jan 12 '17 at 23:51
Namaste
1
asked Jan 12 '17 at 23:45
user406484user406484
111
edited Jan 12 '17 at 23:51
Namaste
1
asked Jan 12 '17 at 23:45
user406484user406484
111
edited Jan 12 '17 at 23:51
Namaste
1
edited Jan 12 '17 at 23:51
Namaste
1
edited Jan 12 '17 at 23:51
Namaste
1
1
asked Jan 12 '17 at 23:45
user406484user406484
111
asked Jan 12 '17 at 23:45
user406484user406484
111
asked Jan 12 '17 at 23:45
user406484user406484
111
111
$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46
$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50
add a comment |
$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46
$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50
$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46
$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46
$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50
$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
$$37^2x^2-36times37x-7times37=0 .$$
To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
$$(37x-18)^2=7times37+18^2 .$$
I'm sure you can finish it from here.
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
$endgroup$
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
add a comment |
$begingroup$
Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.
There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are
Given $ax^2+bx+c=0$,
- Note the values of $4a=148$ and $b^2=1296$
- Subtract $c$ from both sides: $37x^2-36x=7$
- Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$
- Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$
- The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$
From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
$endgroup$
add a comment |
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2 Answers
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$begingroup$
To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
$$37^2x^2-36times37x-7times37=0 .$$
To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
$$(37x-18)^2=7times37+18^2 .$$
I'm sure you can finish it from here.
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
$endgroup$
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
add a comment |
$begingroup$
To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
$$37^2x^2-36times37x-7times37=0 .$$
To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
$$(37x-18)^2=7times37+18^2 .$$
I'm sure you can finish it from here.
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
$endgroup$
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
add a comment |
$begingroup$
To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
$$37^2x^2-36times37x-7times37=0 .$$
To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
$$(37x-18)^2=7times37+18^2 .$$
I'm sure you can finish it from here.
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
$endgroup$
To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
$$37^2x^2-36times37x-7times37=0 .$$
To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
$$(37x-18)^2=7times37+18^2 .$$
I'm sure you can finish it from here.
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
answered Jan 12 '17 at 23:55
DavidDavid
69.4k667130
69.4k667130
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
add a comment |
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format
$endgroup$
– user406484
Jan 13 '17 at 0:00
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
$endgroup$
– user406484
Jan 13 '17 at 0:06
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
The formula you gave in your comment is missing a square but I assume that's just a typo.
$endgroup$
– David
Jan 13 '17 at 0:09
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
$begingroup$
Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
$endgroup$
– David
Jan 13 '17 at 0:10
add a comment |
$begingroup$
Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.
There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are
Given $ax^2+bx+c=0$,
- Note the values of $4a=148$ and $b^2=1296$
- Subtract $c$ from both sides: $37x^2-36x=7$
- Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$
- Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$
- The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$
From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
$endgroup$
add a comment |
$begingroup$
Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.
There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are
Given $ax^2+bx+c=0$,
- Note the values of $4a=148$ and $b^2=1296$
- Subtract $c$ from both sides: $37x^2-36x=7$
- Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$
- Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$
- The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$
From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
$endgroup$
add a comment |
$begingroup$
Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.
There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are
Given $ax^2+bx+c=0$,
- Note the values of $4a=148$ and $b^2=1296$
- Subtract $c$ from both sides: $37x^2-36x=7$
- Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$
- Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$
- The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$
From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
$endgroup$
Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.
There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are
Given $ax^2+bx+c=0$,
- Note the values of $4a=148$ and $b^2=1296$
- Subtract $c$ from both sides: $37x^2-36x=7$
- Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$
- Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$
- The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$
From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
edited Jan 13 '17 at 4:26
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
answered Jan 13 '17 at 1:59
John Wayland BalesJohn Wayland Bales
14.6k21238
14.6k21238
add a comment |
add a comment |
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$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46
$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50