Thread: Finding the Value of x^2-y^2

1. Finding the Value of x^2-y^2

I have this problem here which I can't seem to solve (why is my intro always like this):

$\sqrt{(x^2+y^2-20)^2}+[x^2+y^2+xy-18]=0$

The expression inside the [ ] means the absolute value. I can't seem to express absolute values, so.

When real number x and y satisfies the above equation, what's the value of
$x^2-y^2$?
$(x < y < 0)$

Can anyone help me out here?

2. Originally Posted by lanierms
I have this problem here which I can't seem to solve (why is my intro always like this):

$\sqrt{(x^2+y^2-20)^2}+[x^2+y^2+xy-18]=0$

The expression inside the [ ] means the absolute value. I can't seem to express absolute values, so.

When real number x and y satisfies the above equation, what's the value of
$x^2-y^2$?
$(x < y < 0)$

Can anyone help me out here?
1. Both summands are positive. Thus this equation can only be true if both summands equal zero.

2. From $x^2+y^2-20=0$ follows $x^2+y^2=20$. Plug in this value into the 2nd summand:

$x^2+y^2+xy-18=0~\implies~20+xy-18=0~\implies~y=-\frac2x$

3. The first equation becomes:

$x^2+\frac4{x^2}-20=0~\implies~x^4-20x^2+4=0$

Use the substitution $z = x^2~\implies~|x|=\sqrt{z}$

Solve for z:

$z^2-20z+4=0~\implies~z=10\pm\sqrt{96}$

4. Therefore $|x|=\sqrt{10\pm\sqrt{96}}$. Plug in the appropriate values into the equation at #2.

3. $\sqrt{(x^2+y^2-20)^2}+ |x^2+y^2+xy-15| = 0$

$\sqrt{(x^2+y^2-20)^2} = - (|x^2+y^2+xy-15|)$

$(x^2+y^2-20)^2 = (|x^2+y^2+xy-15|)^2$

$(x^2+y^2-20)^2 -(x^2+y^2+xy-15)^2= 0$

$[(x^2+y^2-20) + (x^2+y^2+xy-15)][(x^2+y^2-20) - (x^2+y^2+xy-15)] = 0$

$[2x^2 + 2y^2 + xy -35][-20 -xy +15] = 0$

[-20 -xy +15] = 0 Or xy = -5

[2x^2 + 2y^2 + xy -35] = 0. Put xy = -5

[2x^2 + 2y^2 -5 -35] = 0 Or

$x^2 + y^2 = 20$

$(x^2 - y^2)^2 = (x^2+y^2)^2 - 4x^2y^2$

Substitute the values and solve for $(x^2 - y^2)$

4. $\sqrt{(x^2+y^2-20)^2}+ |x^2+y^2+xy-15| = 0$

$\sqrt{(x^2+y^2-20)^2} = - (|x^2+y^2+xy-15|)$

$(x^2+y^2-20)^2 = (|x^2+y^2+xy-15|)^2$

$(x^2+y^2-20)^2 -(x^2+y^2+xy-15)^2= 0$

$[(x^2+y^2-20) + (x^2+y^2+xy-15)][(x^2+y^2-20) - (x^2+y^2+xy-15)] = 0$

$[2x^2 + 2y^2 + xy -35][-20 -xy +15] = 0$

[-20 -xy +15] = 0 Or xy = -5

$[2x^2 + 2y^2 + xy -35]$= 0. Put xy = -5

$[2x^2 + 2y^2 -5 -35]$= 0 Or

$x^2 + y^2 = 20$

$(x^2 - y^2)^2 = (x^2+y^2)^2 - 4x^2y^2$

Substitute the values and solve for $(x^2 - y^2)$

5. Thank you everyone for helping out my problem here.
earboth's solution was the easiest to understand.
thank you.
oh yeah, how do you express the absolute value using Latex?
Sorry for putting this topic under Algebra. I didn't know this was Pre-Calculus.

6. Originally Posted by lanierms
oh yeah, how do you express the absolute value using Latex?
On my keyboard, the key is right above "enter" on the same key as backslash (\); so it's shift-backslash.

7. Originally Posted by lanierms
Thank you everyone for helping out my problem here.
earboth's solution was the easiest to understand.
thank you.
Originally Posted by lanierms
...

When real number x and y satisfies the above equation, what's the value of
$\bold{\color{blue}x^2-y^2}$
?
$(x < y < 0)$

Can anyone help me out here?
Originally Posted by sa-ri-ga-ma
$\sqrt{(x^2+y^2-20)^2}+ |x^2+y^2+xy-\bold{\color{red}15}| = 0$

...

If you use
$x^2+y^2=20~\wedge~xy=-2$

then you'll get:

$x^2+y^2=\sqrt{20^2-4 \cdot 4}= 8\cdot \sqrt{6}$

8. Originally Posted by earboth

If you use
$x^2+y^2=20~\wedge~xy=-2$

then you'll get:

$x^2+y^2=\sqrt{20^2-4 \cdot 4}= 8\cdot \sqrt{6}$
Thank you earboth. In that case, the equation

becomes
$(2x^2 + 2y^2 - 38)(20 - xy + 18) = 0$

So

$x^2 + y^2 = 19; and ;xy = -2$

$x^2 - y^2 = \sqrt{19^2 - 4\cdot4)}$

9. Originally Posted by sa-ri-ga-ma
Thank you earboth. In that case, the equation

becomes
$(2x^2 + 2y^2 - 38)(20 - xy + 18) = 0$

So

$x^2 + y^2 = 19; and ;xy = -2$

$x^2 - y^2 = \sqrt{19^2 - 4\cdot4)}$
Next step of the endless story:

In that case the equation:

$\left((8x^2+y^2-20)+(x^2+y^2+xy-18)\right)$ $\left((8x^2+y^2-20) -(x^2+y^2+xy-18)\right) =0$

becomes:

$(2x^2+2y^2\bold{+xy}-38)(20+xy-18)=0$

If the the 2nd bracket equals zero then $xy = -2$.

Therefore the 1st bracket is:

$(2x^2+2y^2-2-38)=0~\implies~x^2+y^2=20$

etc. andsoon ...

10. Originally Posted by earboth
Next step of the endless story:

In that case the equation:

$\left((8x^2+y^2-20)+(x^2+y^2+xy-18)\right)\left((8x^2+y^2-20) -(x^2+y^2+xy-18)\right) =0$

becomes:

$(2x^2+2y^2\bold{\color{blue}+xy}-38)(20+xy-18)=0$

If the the 2nd bracket equals zero then $xy = -2$.

Therefore the 1st bracket is:

$(2x^2+2y^2-2-38)=0~\implies~x^2+y^2=20$

etc. andsoon ...
Oooooo No! Thanks.