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Let $a$ and $b$ be the roots of the equation: $x^2 - 10cx - 11d = 0$ where $c$ and $d$ be the roots of $x^2 - 10ax - 11b = 0$. Find the value of $a+b+c+d$, assuming that they all are distinct.

I first tried making an equation with roots $(a+b)$ and $(c+d)$ to get the sum of the roots, however I wasn't able to solve this question using that method as the answer which I got was in terms of the variables itself.

I also tried placing $a$ into the first equation and $c$ into the second to cancel out a common term ($-10ac$), but after cancelling, I got: $(a^2 - c^2 - 11d + 11b = 0)$. Now I don't know how to move ahead.

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Have a look at this web page, a few lines down, under the heading of quadratic. It goes a long way to solving your problem, I think. mathsisfun.com/algebra/polynomials-sums-products-roots.html –  Blakes7 Jul 16 at 5:36

2 Answers 2

up vote 7 down vote accepted

My Solution ::

Given $a,b$ are the roots of $x^2-10cx-11d=0.$
So $$ a+b = 10c \tag{1}$$
and $$ab = -11d \tag{2} $$

and $c,d$ are the roots of $x^2-10ax-11b=0.$ So $$c+d=10a\tag{3}$$
and $$cd=-11b \tag{4}$$

So $$a+b+c+d = 10(a+c). \tag{5}$$

Now $$\frac{a+b}{c+d} = \frac{10c}{10a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow a^2-11d=c^2-11b$$

So $$(a^2-c^2)=-11(b-d)\Rightarrow (a+c)\cdot(a-c)=-11(b-d) \tag{6}$$

Now $(1)-(3)$, we get $$a+b-c-d=10c-10a\Rightarrow (b-d) = 11(c-a)=-11(a-c)$$

Now putting $(b-d) = -11(a-c)$ in eqn. $(6)$, we get $$(a+c)\cdot(a-c)=121(a-c)$$

So $(a-c)\cdot(a+c-121) = 0$. Now $a\neq c$, because $a,b,c,d$ are distinct real numbers. So $a+c=121$. Put into eqn. $(5)$. We get $$ a+b+c+d = 10(a+c) = 10\cdot 121 = 1210\Rightarrow (a+b+c+d) = 1210. $$

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Thanks a lot! :) –  Harshal Gajjar Jun 15 at 13:22
    
That last line should start with an "a" not a "1", I believe. –  Blakes7 Jun 16 at 11:51

We have

$a+b=10c$ and $c+d=10a$, hence

$$b=10c-a$$ and $$d=10a-c$$

We also have $ab=-11d$ and $cd=-11b$, hence

$$10ac-a^2=-11(10a-c)$$ and $$10ac-c^2=-11(10c-a)$$ Subtracting these gives $$a^2-c^2=-11(10c-a)+11(10a-c)=121(a-c)$$ Assuming $a\not=c$, we get $$a+c=121$$ Going back to the formulas for $b$ and $d$, we have $$b+d=(10c-a)+(10a-c)=9(a+c)$$ Thus $$a+b+c+d=(a+c)+(b+d)=(a+c)+9(a+c)=10(a+c)=1210$$

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