Shemur

Question

Shemur

preetha · Answer

@Sam or @abb0t may be able to help.

abb0t · Answer

What you have here is a system of equations. 
Notice what they have given you: $y=-x^2$

What that means is that you can substitute \)-x^2\) for the $y$ in the first equation. 

It might help if I give them numbers

A.) $x^2 - (y - 12)^2 = 144$
B.) $y=-x^2$

What I meant by substitution was this, plug in the y function into $A$ wherever you see a $y$

So, $x^2-(\sf\color{#8C00FF }{-x^2}-12)^2=144$
now, solve for $x$ 

Can you finish this now?

abb0t · Answer

Once you have (if you have a solution), plug the given value of $x$ into equation $B$ and you should get a value for $y$

abb0t · Answer

You need to perform the distributive operation: $(-x^2-12)(-x-12)$

abb0t · Answer

$(-x^2-12)(-x^2-12)$*

phi · Answer

after expanding the square you get
$$ x^2 - (-x^2 - 12)^2 = 144 \ x^2 -(x^4 +24x^2+144)= 144\ -x^4 -23x^2 -288= 0 \ x^4 +23x^2 +288=0 $$
if you let u = x^2, this is
$$ u^2 +23u + 288= 0$$
but you will have to take the square root of a complex number to get numerical values