qaudratic eqautions

**ENNO** · February 25th 2008, 09:40 AM

X-2=10
X(SQAURED)+Y(SQAURED)=20

I cant seem to get the formula to work...keeps on coming out uneven!!!

**Jhevon** · February 25th 2008, 09:43 AM

Originally Posted by ENNO

X-2=10
X(SQAURED)+Y(SQAURED)=20

I cant seem to get the formula to work...keeps on coming out uneven!!!

check your question again. it makes no sense to write x - 2 = 10. why not just write x = 12?

**ENNO** · February 25th 2008, 09:48 AM

Originally Posted by Jhevon

check your question again. it makes no sense to write x - 2y = 10. why not just write x = 12?

because I just took it straight from the book...this doesnt help my problem in any way!

**Jhevon** · February 25th 2008, 09:51 AM

Originally Posted by ENNO

because I just took it straight from the book...this doesnt help my problem in any way!

ok. i think you need to provide more info. because i don't see what you want to do here. it seems you want to solve the equations simultaneously, but then the solution for x is given in the first equation which makes the problem trivial. simply substitute x = 12 into the second equation to solve for y. the problem is stupid, unless i am misinterpreting the directions and/or you made a typo

**ENNO** · February 25th 2008, 09:55 AM

Originally Posted by Jhevon

ok. i think you need to provide more info. because i don't see what you want to do here. it seems you want to solve the equations simultaneously, but then the solution for x is given in the first equation which makes the problem trivial. simply substitute x = 12 into the second equation to solve for y. the problem is stupid, unless i am misinterpreting the directions and/or you made a typo

yes I did...it is a simultaneous eqaution...sorry bout that!

And alos it is a 2y...

**Jhevon** · February 25th 2008, 09:59 AM

Originally Posted by ENNO

yes I did...it is a simultaneous eqaution...sorry bout that!

And alos it is a 2y...

so you did make a typo!

well, anyway. keep the same game plan. from the first equation then, x = 10 + 2y

plug this in for x in the second equation and solve for y. then you can find x

**ENNO** · February 25th 2008, 10:02 AM

Originally Posted by Jhevon

so you did make a typo!

well, anyway. keep the same game plan. from the first equation then, x = 10 + 2y

plug this in for x in the second equation and solve for y. then you can find x

Thats my problem,

I can work it out into a qaudratic formula but it doesnt work out!

80-40y-5y(sqaured)=0....when I plug it into the qaudratic formula i get crazy answers!

**Jhevon** · February 25th 2008, 10:08 AM

Originally Posted by ENNO

Thats my problem,

I can work it out into a qaudratic formula but it doesnt work out!

80-40y-5y(sqaured)=0....when I plug it into the qaudratic formula i get crazy answers!

ok, we have:

$x - 2y = 10$ .................(1)
$x^2 + y^2 = 20$ ...........(2)

From (1), $x = 10 + 2y$ , plug this into (2), we get:

$(10 + 2y)^2 + y^2 = 20$

$\Rightarrow 100 + 40y + 4y^2 + y^2 = 20$

$\Rightarrow 5y^2 + 40y + 80 = 0$

$\Rightarrow y^2 + 8y + 16 = 0$

this is a perfect square, we don't even need the quadratic formula here

$\Rightarrow (y + 4)^2 = 0$

$\Rightarrow y = -4$

now continue

**ENNO** · February 25th 2008, 10:28 AM

Originally Posted by Jhevon

ok, we have:

$x - 2y = 10$ .................(1)
$x^2 + y^2 = 20$ ...........(2)

From (1), $x = 10 + 2y$ , plug this into (2), we get:

$(10 + 2y)^2 + y^2 = 20$

$\Rightarrow 100 + 40y + 4y^2 + y^2 = 20$

$\Rightarrow 5y^2 + 40y + 80 = 0$

$\Rightarrow y^2 + 8y + 16 = 0$

this is a perfect square, we don't even need the quadratic formula here

$\Rightarrow (y + 4)^2 = 0$

$\Rightarrow y = -4$

now continue

I worked it out and got X=18...am I right?

**Jhevon** · February 25th 2008, 07:53 PM

Originally Posted by ENNO

I worked it out and got X=18...am I right?

no

**mr fantastic** · February 25th 2008, 08:01 PM

Originally Posted by ENNO

I worked it out and got X=18...am I right?

You're nearly right ....... It's not 10 plus 8 .....

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