Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $c=2a-d$.

share|cite|improve this question
Have you considered the products of the roots,ab=-5d and cd=-5b,solving for 'a' – DOCTOR NGILAZI BANDA JOSHUA Jul 7 '13 at 23:56
up vote 2 down vote accepted

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

share|cite|improve this answer
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 '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 $x$.

share|cite|improve this answer
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

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.

share|cite|improve this answer

$\bf{My\; Solution::}$ 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 $a+b+c+d = 2(a+c) = 2\cdot 15 = 30\Rightarrow \boxed{\boxed{a+b+c+d = 30}}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.