Question

If the roots of the equation $x^2-10cx-11d=0$ are $a,b$ and those of $x^2-10ax-11b=0$ are $c,d$, then the positive value of $a+b+c+d$ is $(a,b,c$ and $d$ being distinct numbers $)$ is $1210$

• A. True
• B. False

Answer 1

The correct option is A True

Here, $a+b=10c$ and $c+d=10a$
$\Rightarrow (a-c)+(b-d)=10(c-a)\Rightarrow (b-d)=11(c-a)$ ...(i)
Since, ${}^{\prime }{c}^{\prime }$ is the root of

$x^2-10ax=10b=0$$\Rightarrow c^2-10ac=11b=0$ ...(ii)
Similarly, ${}^{\prime }{a}^{\prime }$ is the root of
$x^2-10cx-11d=0\Rightarrow a^2-10ca-11d=0$ ...(iii)
On subtracting Eq. (iv) from Eq. (ii), we get
$(c^2-a^2)=11(b-d)$ ...(iv)
$\therefore (c+a)(c-a)=11\times 11(c-a)$ [from Eq.(i)]
$\Rightarrow c+a=121$
$\therefore a+b+c+d=10c+10a=10(c+a)=1210$