If $x^2-10ax-11b=0$ have roots $c$ and $d$. $x^2 -10cx-11d=0$ have roots $a$ and $b$ then the value of $a+b+c+d$ is..
Here from both equations, $c+d =10a$ and $a+b = 10c $ and $ac =121$. But how I can get the sum $a+b+c+d$ from these!
Please help. Thanks in advance.
Given $$x^2-10ax-11b=0\tag{1}$$ $$x^2-10cx-11d=0\tag{2}$$ We know that $c,d$ are roots of $(1)$ and that $a,b$ are roots of $(2)$ so: $$\begin{cases}{c^2 - 10ac - 11b = 0\\a^2 - 10ac - 11d = 0}\end{cases}$$ Subtracting then we obtain:
$$a^2 - c^2 = 11 ( d - b )\tag{3}$$ Then $$a + b = 10c$$ $$ab = - 11d$$ $$c + d = 10a$$ $$cd = - 11b$$ And we get that $$a + b + c + d = 10 ( a + c )\tag{4}$$ $$b + d = 9 ( a + c ) \tag{5}$$ $$ac=121\tag{6}$$ Using $(3)$ and $(5)$ we get: $$\frac{( b + d ) ( a - c )}9 = 11 ( d - b )$$ $$\frac{ a - c }{99} = \frac{ d - b }{ b + d }$$ Then $$a^2 + c^2 - 20 ac - 11 ( b + d ) = 0$$ Using $(6)$ and solving for $a+c$ we have: $$( a + c )^2 - 22ac - 11 ( a + c ) 9 = 0\implies a+c=121\tag{7}$$ Finally from $(4)$ and $(7)$ we obtain $$\color{orange}{a+b+c+d=1210}$$