Question

If $x^2-10ax-11b=0$ has the roots $c$ and $d$, then $x^2-10x-11d=0$ has the roots $a$ and $b$. Find the value of $a+b+c+d$ if $a+c$ is positive.

• A. $1220$
• B. $1110$
• C. $1210$
• D. $1310$

Answer 1

The correct option is C $1210$

Given: $c,d$ are roots of $x^2-10ax-11b=0$

$\therefore c+d=10a$ ...... (1)

$cd=-11b$ ...... (2)

Now $\left(1\right)\times c$ and using equation (2) we get,

$c^2-10ac-11b=0.......\left(3\right)$

Also, $a,b$ are roots of $x^2-10cx-11d=0$

$a+b=10c$.... (4)

$ab=-11d$ .... (5)

Multiplying the equation(4) by $a$ and using the 5th equation we get,

$a^2-10ac-11d=0.......\left(6\right)$

Now from the above equations we have,

$a+b+c+d=10(a+c)\Rightarrow b+d=9(a+c)$ ..... (7)

$abcd=121bd$

$\therefore ac=121$

Now, $eq\left(3\right)-eq.\left(6\right)\Rightarrow (c^2-a^2)-11(b-d)=0$

$c^2-a^2=11(b-d)$

$eq.\left(1\right)+eq.\left(2\right)\Rightarrow a^2+c^2-20ac-11(b+d)=0$

Using the identity $(a+b{)}^2=a^2+b^2+2ab$ and eq(7) in the above eq. we get,

${(a+c)}^2-22ac-99(a+c)=0$

${(a+c)}^2-99(a+c)-2662=0$ [$\because ac=121$]

${(a+c)}^2-(121-22)(a+c)-121\times 22=0$

$(a+c{)}^2-121(a+c)+22(a+c)-121\times 22=0$

$(a+c)(a+c-121)+22(a+c-121)=0$

$(a+c+22)(a+c-121)=0$

$\therefore a+c=121$

$a+b+c+d=10(a+c)$

$\therefore a+b+c+d=1210$