Power pack

Algebra Level 4

1 6 x 2 + y + 1 6 x + y 2 = 1 \large 16^{x^{2}+y}+16^{x+y^{2}}=1 Given that x x and y y are real numbers that satisfy the equation above. Determine the number of ordered pairs of ( x , y ) (x,y) .


The answer is 1.

View wiki
Dhiraj Agarwalla by dhiraj agarwalla , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Dhiraj Agarwalla
Dec 3, 2015

Applying A.M>=G.M 1 6 x 2 + y + 1 6 x + y 2 2 > = 1 6 x 2 + y ∗ 1 6 x + y 2 \frac{16^{x^{2}+y}+16^{x+y^{2}}}{2}>=\sqrt{16^{x^{2}+y}*16^{x+y^{2}}} 1 6 x 2 + y + 1 6 x + y 2 2 > = 1 6 x 2 + x + y 2 + y \frac{16^{x^{2}+y}+16^{x+y^{2}}}{2}>=\sqrt{16^{x^{2}+x+y^{2}+y}} x 2 + x > = − 1 4 x^{2}+x>= -\frac{1}{4} 1 6 x 2 + y + 1 6 x + y 2 2 > = 1 6 − 1 2 \frac{16^{x^{2}+y}+16^{x+y^{2}}}{2}>=\sqrt{16^{-\frac{1}{2}}} 1 6 x 2 + y + 1 6 x + y 2 > = 1 16^{x^{2}+y}+16^{x+y^{2}}>=1 The above condition is fulfilled only when both the terms of A.M are equal. Thus x=y . Thus the only possible solution is ( − 1 2 , − 1 2 ) (-\frac{1}{2},-\frac{1}{2}) .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...