this is in algebra?

Question

If a ,b are the roots of the equation x2-10cx-11d=0 and c, d are the roots of the equation x2-10ax-11d=0,then a + b +c + d equals (a ,b ,c, d are distinct real numbers)[multiple choice]
(Ans) (A) 1101 (B) 1200 (C) 1210 (D) none of these

the 2 near 'x' in the 2 equations are square

Answer 1

a = (10c + sqrt(100c^2 + 44d))/2 = 5c + sqrt(25c^2 + 11d)
b = (10c - sqrt(100c^2 + 44d))/2 = 5c - sqrt(25c^2 + 11d)
c = (10a + sqrt(100a^2 + 44d))/2 = 5a + sqrt(25a^2 + 11d)
d = (10a - sqrt(100a^2 + 44d))/2 = 5a - sqrt(25a^2 + 11d)

The radicals cancel out and leave you with:

a + b + c + d =
10c/2 + 10c/2 + 10a/2 + 10a/2
= 10c + 10a

We can't figure it out without more information.

Answer 2

I tried to solve it, and i am not too sure how to solve it algebraically.

But something i did notice, if a, b, c, and d are all zero, that would make them all roots of the equations (the equations would be x^2 = 0 ), and then a + b + c + d = 0.

Edit: I think you may be able to solve for the values by looking at it, if x^2 - 10cx - 11d = (x + a)(x + b), then a + b = -10c, and a*b = -11d.

You can do the same with the other equation, x^2 - 10ax - 11d = (x + c)(x + d), c + d = -10a, and c*d = -11d, but with the second problem, if you divide both sides by d, then you get c = -11, so whatever other stuff is there, c = -11. you did mean for the second equation to have a 'd' there, and not a 'b', right?

After that, you can possibly solve them for a, b, and d.

When i did that, i came out with a = 5.5(9 +/- sqrt(85) ), b = 110 - a, c = -11, and d = 11-10a, but i am not too sure how accurate that is. when i factored the first equation with those values, it was off by a little bit. But even if you do figure them out, i am fairly certain that it will not be any of the three numerical options on the multiple choice, so it is D. None of these.

Answer 3

Since a ,b are the roots of the equation x^2-10cx-11d=0
we have a^2 - 10ac - 11d =0 and c, d are the roots of the equation x2-10ax-11d=0
we have c^2 - 10ac - 11d =0
then solving these two equations we get, a^2 - c^2 = 0
=> (a+c)(a-c) = 0 => a = +/- c & similarly, we get b = +/- d.
And one more condition we have all the four roots are distinct real numbers.
So, it should be +a, -a, +c & -c

Now, a+b+c+d = 0
So the answer should be (D) none of these

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Answer 4

1210

Answer 5

(D) none of these

x2-10ax-11d = (x-c)(x-d)
-10a = -c-d => c+d = 10a
-11d = cd => -11 = c

x2-10cx-11d = (x-a)(x-b)
-10c = -b-a => a+b = 10c
-11d = ab
-11(10a-c) = a(10c-a)
-110a + 11c = 10ac - a^2
-110a + 11(-11) = 10a(-11) - a^2
a^2 = 121
a = 11, -11

a+b+c+d = 10c + 10a = 10(-11) + 10(11) = 0
a+b+c+d = 10c + 10a = 10(-11) + 10(-11) = -220

Answer 6

from the eqns

a+b=10c
c+d=10a
ab= -11d
cd= -11d

c= -11
a+b+c=11c

now we have to find value of d
b=10c-a
d=10a-c

ab=cd= -11
a(10c-a)=c(10a-c)
a2=c2
a=+/-c

if a= -c
d= -11c
a+b+c+d=0

if a=c

d=9c

a+b+c+d=20c
=20(-11)
=-220

hence
tick
none of these

Answer 7

Of course this is algebra... I don't know if it makes any sence but yes it is because it uses variables.

Answer 8

Yep, that's algebra.... gotta love it. lol Unfortunately, that is confusing.