The expression 3x^2 + 2xy − 8y^2 − 8x + 14y −3 can be treated as a function in x with y as an arbitrary constant or as a function in y with x as an arbitrary constant. Show that in either case the expression is indefinite. Factor the expression.

Question

anonymous · Answer

@AonZ By indefinite is the question saying that in either case the function approaches infinity as x or y approach infinity?

aonz · Answer

i dont think so. Indefinite means you that 
$$\Delta \ge0$$

unklerhaukus · Answer

$$\Delta=b^2-4ac$$?

unklerhaukus · Answer

if y is a constant 

$$(3)x^2 + (2y-8)x +( 14y- 8y^2 −3)$$
$$(a)x^2+(b)x+(c)$$

aonz · Answer

yea i done that then i got 25y^2 - 34y + 7 
when i used the discriminant

anonymous · Answer

With the same approach, if 'x' is constant then we get:$$\bf (-8)y^2+(14+2x)y+(3x^2-8x-3)$$$$\bf -8y^2+(d)y+(e)$$

anonymous · Answer

@AonZ So $\bf \Delta$ does represent the discriminant?

aonz · Answer

yes

anonymous · Answer

Ok so in each case, we'll have to consider the discriminant separately. I'll first consider the case in which 'x' is considered a constant. In such a case, the discriminant is:$$\bf b^2-4ac=(14+2x)^2-4(-8)(3x^2-8x-3)$$$$\bf =196+56x+4x^2+96x^2-256x-96=100x^2-200x+100$$$$\bf =100(x^2-2x+1)=100(x-1)^2= \Delta \ge 0$$So we have proved the discriminant is always greater than 0 when we consider 'x' as a constant. Now you must do the same when we consider 'y' a constant. 

@AonZ

aonz · Answer

oh true...  then after we proved it is always greater of less than 0 we factorise the expression?

anonymous · Answer

I believe that for the factoring part they are asking you to factor the original equation that is given..

aonz · Answer

ok, that shouldnt be too hard. Thanks for your help!!!

anonymous · Answer

yw