MHB Proving No Common Solution for Equations (1) and (2)

[Albert1]

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

 

[HallsofIvy]

Spoiler

Looks straightforward. If there exist a number, x, such that
$x^2- abx+ (a+ b)= 0$ and
$x^2- (a+ b)x+ ab= 0$

then, subtracting, $(a- ab+ b)x+ (a- ab+ b)= 0$ which has the single solution $x= -1$. Putting $x= -1$ into both equations, $1+ ab+ a+ b= 0$ and $1+ a+ b+ ab= 0$. Since a and b are both positive numbers, that is impossible.

 

[kaliprasad]

Spoiler

Looks straightforward. If there exist a number, x, such that
$x^2- abx+ (a+ b)= 0$ and
$x^2- (a+ b)x+ ab= 0$

then, subtracting, $(a- ab+ b)x+ (a- ab+ b)= 0$ which has the single solution $x= -1$. Putting $x= -1$ into both equations, $1+ ab+ a+ b= 0$ and $1+ a+ b+ ab= 0$. Since a and b are both positive numbers, that is impossible.

Spoiler

One part was overlooked that a+b - ab = 0 or (a-1)(b-1) =1 this is pssible for positive a, b ( for example a = 1.5, b= 3) but not for given condition and b both greater than 2 as both terms are geater han 1 so the product.