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 $c=2a-d$.

  • 2
    $\begingroup$ Have you considered the products of the roots,ab=-5d and cd=-5b,solving for 'a' $\endgroup$ – DOCTOR NGILAZI BANDA JOSHUA Jul 7 '13 at 23:56

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 \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.

  • $\begingroup$ 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. $\endgroup$ – Mick Dec 3 '13 at 4:48
  • $\begingroup$ @Mick, thank you for accurate reading. I fixed typo now. $\endgroup$ – Oleg567 Jan 9 '14 at 12:17

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 $x$.

  • $\begingroup$ how to solve these 4 equations $\endgroup$ – maths lover Jul 8 '13 at 3:05
  • $\begingroup$ @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. $\endgroup$ – augurar Jul 13 '13 at 7:25

$\bf{My\; Solution::}$ Given $a\;,b$ are the roots of the equation $x^2-2cx-5d=0$ So

$\displaystyle a+b=2c............................(1)\;\;\;\;\;\; ab = -5d.....................(2)$

similarly $c\;,d$ are the roots of the equation $x^2-2ax-5b=0$ So

$\displaystyle c+d=2a............................(3)\;\;\;\;\;\; cd=-5b.....................(4)$

So $a+b+c+d = 2(a+c)............(5)$

Now $\displaystyle \frac{a+b}{c+d}=\frac{2c}{2a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow (a^2-c^2)=(cd-ab)=-5(b-d)$

So $\displaystyle (a+c)\cdot (a-c)=-5(b-d)=-5\left\{(2c-a)-(2a-c)\right\}=-15\left\{c-a\right\}=15(a-c)$

So $\displaystyle (a+c)\cdot (a-c)-15(a-c)=0\Rightarrow (a-c) = 0$ or $(a+c) = 15$

Now $a\neq c\;,$ bcz $a,b,c,d$ are distinct real no.

So $a+c-15=0\Rightarrow a+c = 15$. put into eqn....$(5)$

We get $a+b+c+d = 2(a+c) = 2\cdot 15 = 30\Rightarrow \boxed{\boxed{a+b+c+d = 30}}$


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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