- #1
Albert1
- 1,221
- 0
$ assume \,\, a>2\,\, and \,\, b>2$
$x^2-abx+(a+b)=0---(1)$
$x^2-(a+b)x+ab=0---(2)$
$prove \,\, (1)\,\, and \,\, (2)\,\, have \,\, no\,\ common \,\, solution$
$x^2-abx+(a+b)=0---(1)$
$x^2-(a+b)x+ab=0---(2)$
$prove \,\, (1)\,\, and \,\, (2)\,\, have \,\, no\,\ common \,\, solution$