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Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then find the value of $a+b+c+d$.

I could only get $2$ equations that

$a=2c-b$ and $c=2a-d$.

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Have you considered the products of the roots,ab=-5d and cd=-5b,solving for 'a' –  JOSHUA BANDA Jul 7 '13 at 23:56
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3 Answers 3

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Vieta's formulas $\Rightarrow \ \ $ $a+b=2c$, $c+d=2a$ $\ \ \Rightarrow \ \ $ $a+b+c+d=2(a+c)$ $\ \Rightarrow \ $ $a+c=b+d$.

Denote $$m = \dfrac{a+c}{2}=\dfrac{b+d}{2}, \qquad p=\dfrac{c-a}{2}\color{gray}{=m-a=c-m}.$$

Then $$ \left\{ \begin{array}{r} a = m-p, \quad b=m+3p, \\ c = m+p, \quad d=m-3p. \end{array} \right. $$

Vieta's formulas $\Rightarrow \ $ $ab=-5d$, $\ \ $ $cd=-5b$ $\ \ \Rightarrow$ $$ \left\{ \begin{array}{r} (m-p)(m+3p)=-5m+15p, \\ (m+p)(m-3p)=-5m-15p; \end{array} \right. $$ $$ \left\{ \begin{array}{r} m^2+2mp-3p^2=-5m+15p, \\ m^2-2mp-3p^2=-5m-15p; \end{array} \right. $$ $$ \left\{ \begin{array}{c} m^2-3p^2=-5m, \\ 2mp=15p. \end{array} \right. $$ Since $a,b,c,d$ are distinct, then $p\ne 0$, then $2m=15$, then $$\color{#660011}{\Large{a+b+c+d=4m=30}}.$$

Note: $3p^2=m^2+5m=\dfrac{375}{4}$ $\ \ \Rightarrow \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.

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Could there be a typo? At first, you stated p = a - m. Then at a later stage, you said a = m - p. The two statements seem to be contradictory. The same thing happened to p = c + m. –  Mick Dec 3 '13 at 4:48
    
@Mick, thank you for accurate reading. I fixed typo now. –  Oleg567 Jan 9 at 12:17
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Let the two polynomials be $$ p(x) = x^2 - 2cx -5d \\ q(x) = x^2 - 2ax - 5b $$ You also know that $$ p(x) = (x-a)(x-b) = x^2 - (a+b)x + ab \\ q(x) = (x-c)(x-d) = x^2 - (c+d)x + cd $$

You might try playing around with these two forms. For example, you can take the product of the polynomials and equate the coefficients for each power of $x$.

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how to solve these 4 equations –  maths lover Jul 8 '13 at 3:05
    
@mathslover Oleg567's answer picks up where this one leaves off. First equate the two forms of each polynomial and use the fact that the coefficients are equal. That gives you $2cx=a+b$, $-5d=ab$, $2ax=c+d$, and $5b=cd$. (I guess these are called Vieta's formulas?) Then use the symmetry of the problem to reduce the number of variables and solve using algebra. –  augurar Jul 13 '13 at 7:25
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For a hint, try plugging in $a$ and $b$ into your first equation since you know that they make the equality true. That will get you two equations. You can do the same thing with $c$ and $d$ in the second equation.

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