using substitution to find coordinates of two quadratic curves, problems

**chartsy** · November 7th 2010, 08:02 AM

i cannot understand where i am going wrong, the question is to find where the two curves
meet:
(squared=

)

x

+y

=25 (1)
y=x

-5 (2)

heres how i worked it out:
(1) y

=25-x

y=5-x
subst y into 2
5-x=x

-5
x

+x-10=0
this does not factorise! help! i am mathematically challenegd and i work hard to try abd get good grades and things like this frustrate me, i can do all the other simluatneous eqns except this type

**skeeter** · November 7th 2010, 08:23 AM

Originally Posted by chartsy

i cannot understand where i am going wrong, the question is to find where the two curves
meet:

next time, use the caret symbol (^) to show an exponent , not an emoticon.

x^2 + y^2 = 25 (1)
y=x^2 - 5 (2)

from equation (2) ...

x^2 = y + 5

substitute y+5 for x^2 in equation (1) ...

(y+5) + y^2 = 25

y^2 + y - 20 = 0

factor, solve for y, then determine the value of x.

also, fyi ... if y^2 = 25 - x^2 , then y does not equal 5 - x

**Archie Meade** · November 7th 2010, 08:24 AM

Originally Posted by chartsy

i cannot understand where i am going wrong, the question is to find where the two curves
meet:
(squared=

)

x

+y

=25 (1)
y=x

-5 (2)

heres how i worked it out:
(1) y

=25-x

y=5-x this is your error

subst y into 2
5-x=x

-5
x

+x-10=0
this does not factorise! help! i am mathematically challenegd and i work hard to try abd get good grades and things like this frustrate me, i can do all the other simluatneous eqns except this type

$x^2+y^2=25=5^2$

$y=x^2-5\Rightarrow\ x^2=y+5$

Now you can make the substitution!

$y^2+y+5=25$

Now you solve the quadratic..

Remember!! $y^2=25-x^2\Rightarrow\ y=\pm\sqrt{25-x^2}$

**chartsy** · November 7th 2010, 09:41 AM

Originally Posted by skeeter

from equation (2) ...

x^2 = y + 5

substitute y+5 for x^2 in equation (1) ...

(y+5) + y^2 = 25

y^2 + y - 20 = 0

factor, solve for y, then determine the value of x.

also, fyi ... if y^2 = 25 - x^2 , then y does not equal 5 - x

?

**skeeter** · November 7th 2010, 09:50 AM

Originally Posted by chartsy

?

what is your problem ? ... can you not factor $y^2 + y - 20 = 0$ ?

**chartsy** · November 7th 2010, 09:50 AM

o, wow i've got to read up

**chartsy** · November 7th 2010, 09:51 AM

Originally Posted by skeeter

what is you problem ? ... can you not factor $y^2 + y - 20 = 0$ ?

i can do that easily it was just what you said the y^2=25-x^2 isn't y=5-x

**Archie Meade** · November 7th 2010, 09:57 AM

$y^2=25-x^2=(5+x)(5-x)$

Only the square root of $(5+x)(5+x)$ is $5+x$ .

For example

$5^2=3^2+4^2$ but $5\ \ne\ 3+4$

You have to take the square root of the entire RHS, not the square roots of each part of it...

$5=\sqrt{3^2+4^2}$

so if $y^2=25-x^2\Rightarrow\ y=\pm\sqrt{25-x^2}$

**skeeter** · November 7th 2010, 09:59 AM

Originally Posted by chartsy

i can do that easily it was just what you said the y^2=25-x^2 isn't y=5-x

what if x = 4 , what would you get for y ?

do both equations give the same result?

**chartsy** · November 7th 2010, 10:06 AM

so if i have to solve two quadratic curves simultaneously i should try and find the value of x/y with the equation which only has one value squared

**chartsy** · November 7th 2010, 10:12 AM

Originally Posted by skeeter

what if x = 4 , what would you get for y ?

do both equations give the same result?

ok, so i will have to take the square root of the whole of the other sidem rather than the individual values

**Archie Meade** · November 7th 2010, 10:35 AM

Originally Posted by chartsy

so if i have to solve two quadratic curves simultaneously i should try and find the value of x/y with the equation which only has one value squared

No, you only need to solve one quadratic.

When you have found y, you can use it's value(s) to find x.

$x^2+y^2=25$

$y=x^2-5\Rightarrow\ x^2=y+5\Rightarrow\ y^2+y+5=25$

You need to solve this for y.
Then you can find x using the y solution(s).

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