Find Real Triples $(a,b,c)$ for 20c-16b^2=9

[anemone]

Find all triples $(a,\,b,\,c)$ of real numbers such that

$20c-16b^2=9$

$24b-36a^2=1$

$12a-4c^2=25$

 

[kaliprasad]

Find all triples $(a,\,b,\,c)$ of real numbers such that

$20c-16b^2=9$

$24b-36a^2=1$

$12a-4c^2=25$

I get

Spoiler

no solution

as

Spoiler

adding all 3 we get

$(6a-1)^2 + (4b-3)^2 + (2c-5)^2=0$

giving $a=\dfrac{1}{6}, b= \dfrac{3}{4}, c=\dfrac{5}{2}$

which shows that above equations inconsistent

 

[anemone]

I get

Spoiler

no solution

as

Spoiler

adding all 3 we get

$(6a-1)^2 + (4b-3)^2 + (2c-5)^2=0$

giving $a=\dfrac{1}{6}, b= \dfrac{3}{4}, c=\dfrac{5}{2}$

which shows that above equations inconsistent

Your solution is correct, and thanks for participating, kaliprasad!