Question for math lovers -1
if root of the equation x 2 − 1 0 c x − 1 1 d = 0 are a,b and those of x 2 − 1 0 a x − 1 1 b = 0 are c,d then find the value of a+b+c+d (where a,b,c,d are distinct numbers)
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3 solutions
You got the wrong answer because your processing of ac = 121 works only if a and c are integers. Unfortunately, the problem does not state whether any of the roots are integers.
Your answer is wrong since there is a substitution
mistake.
1
0
(
a
+
c
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=
1
0
∗
1
2
1
=
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Your method is perfect.
We know if f(x) has roots a,b then f(a)=f(b)=0 so...
a 2 − 1 0 a c − 1 1 d = c 2 − 1 0 a c − 1 1 b
a 2 − 1 1 d = c 2 − 1 1 b
a 2 − c 2 = 1 1 d − 1 1 b
( a + c ) ( a − c ) = 1 1 ( d − b )
( a + c ) = 1 1 a − c d − b
Using sum of roots we can determine that
a + b = 1 0 c
c + d = 1 0 a
using these we can determine that...
( a + b ) + ( c + d ) = 1 0 a + 1 0 c
a + b + c + d = 1 0 ( a + c )
and..
c + d − ( a + b ) = 1 0 a − 1 0 c
d − b = 1 1 ( a − c )
Summarizing, we know now that
a + b + c + d = 1 0 ( a + c )
( a + c ) = 1 1 a − c d − b
d − b = 1 1 ( a − c )
therefore
a + b + c + d = 1 0 ( a + c )
= 1 0 ∗ 1 1 a − c d − b
= 1 1 0 a − c 1 1 ( a − c )
= 1 1 0 ∗ 1 1
= 1 2 1 0
we know that a,b are the roots of the first equation and c,d are the rotes of the second equation
\left{ x²-10cx-11bx=0\ x²-10ax-11bx=0 \right \Longrightarrow -10cx-11dx=-10ax-11bx\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Longrightarrow 10(ax-cx)=11(dx-bx)
thatwhy the value of a+b+c+d must be a multiple of 10 and 11 and the only root how respect this condition is 1210
You can't equal those two equations because they don't have the same roots.
{
x
2
−
1
0
c
x
−
1
1
d
=
0
x
2
−
1
0
a
x
−
1
1
b
=
0
}
⟹
−
1
0
c
x
−
1
1
d
x
=
−
1
0
a
x
−
1
1
b
x
⟹
1
0
(
a
x
−
c
x
)
=
1
1
(
d
x
−
b
x
)
That is why the value of a+b+c+d must be a multiple of 10 and 11 and the only root how respect this condition is 1210
Just used Latex properly. It begins with \ ( and ends with \ ) , ....use \ { .... and... \ }. Note there should be NO space between \ and (,),{,}. I had to keep it for technical reason. In my notes also there is a Latex notes. However please note that your method is not correct as stated by Aldo Culquicondor below.
a+b=10c c+d=10a ab=-11d cd=-11b abcd=121bd which implies ac=121,the factors of 121 are 1,11 and 121 As the numbers are unique,a and c should be 1 and 121 to satisfy the condition. And a+b+c+d=10(a+c) which gives the answer as 10(122)=1220 So I think the given answer is wrong and the answer is not there in the options