JEE Advanced · Mathematics · 2. Quadratic Equations
If \(x^2-10 a x-11 b=0\) have roots \(c\) and \(d\) and \(x^2-10 c x-11 d=0\) have roots \(a\) and \(b\), then, find \(a+b+c+d\).
- A 1200
- B 1210
- C 1220
- D 1320
Answer & Solution
Correct Answer
(B) 1210
Step-by-step Solution
Detailed explanation
Here, \(a+b=10 c\)
\(c+d=10 a\)
On subtracting Eq. (ii) from Eq. (i), we get, \(\quad(a-c)+(b-d)=10(c-a)\) \(\Rightarrow (b-d)\) \(=11(c-a)\)
Since, \(c\) is the root of \(x^2-10 a x+11 b=0\)
\(\Rightarrow c^2-10 a c-11 b=0\)
Similarly, \(a\) is the root of
\(\begin{aligned}\Rightarrow & x^2-10 c x+11 d=0 \\& a^2-10 c a-11 d=0\end{aligned}\)
On subtracting Eq. (v) from Eq. (iv), we get
\(\left(c^2-a^2\right)=11(b-d) \)
\( \therefore (c+a)(c-a)=11 \times 11(c-a)\) \(\{\text {from (i) (ii) } \)
\( \Rightarrow c+a=121 . \)
\( \therefore a+b+c+d=10 c+10 a \)
\( =10(c+a)=1210\)
\(c+d=10 a\)
On subtracting Eq. (ii) from Eq. (i), we get, \(\quad(a-c)+(b-d)=10(c-a)\) \(\Rightarrow (b-d)\) \(=11(c-a)\)
Since, \(c\) is the root of \(x^2-10 a x+11 b=0\)
\(\Rightarrow c^2-10 a c-11 b=0\)
Similarly, \(a\) is the root of
\(\begin{aligned}\Rightarrow & x^2-10 c x+11 d=0 \\& a^2-10 c a-11 d=0\end{aligned}\)
On subtracting Eq. (v) from Eq. (iv), we get
\(\left(c^2-a^2\right)=11(b-d) \)
\( \therefore (c+a)(c-a)=11 \times 11(c-a)\) \(\{\text {from (i) (ii) } \)
\( \Rightarrow c+a=121 . \)
\( \therefore a+b+c+d=10 c+10 a \)
\( =10(c+a)=1210\)
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