1. ## algebraic equation problem

There are 2 pairs of numbers: (a,c) & (b,d).

(a,c) is the root of x^2 + ax - b = 0

(b,d) is the root of x^2 + cx + d = 0

Find all possible real values for a, b, c and d.

2. ## Systems

It's the undertaker. Hi

a and c are the roots of the equation $\displaystyle x^2+ax-b=0$
Then $\displaystyle (x-a)(x-c)=x^2+ax-b$
$\displaystyle x^2-(a+c)x+ac=x^2+ax-b$
Equating coefficients
$\displaystyle -(a+c)=a$ $\displaystyle ac=-b....(1)$
$\displaystyle c=-2a.....(2)$

b and d are the roots of the equation $\displaystyle x^2+cx+d=0$
Then $\displaystyle (x-b)(x-d)=x^2+cx+d$
$\displaystyle x^2-(b+d)+bd=x^2+cx+d$
Equating coefficients
$\displaystyle -(b+d)=c...(3)$ $\displaystyle bd=d.....(4)$

We have a system of 4 equations
$\displaystyle ac=-b....(1)$ $\displaystyle c=-2a.....(2)$
$\displaystyle -(b+d)=c...(3)$ $\displaystyle bd=d.....(4)$

From (4):$\displaystyle bd=d$
$\displaystyle b=1$

Into (3):$\displaystyle -1-d=c$ and into (1): $\displaystyle ac=-1$
(2) into (1):$\displaystyle -2a^2=-1\longrightarrow a^2=1/2$
$\displaystyle a=\pm\sqrt\frac{1}{2}$

Hence c$\displaystyle =-2(\pm\sqrt\frac{1}{2})$

Into (4):$\displaystyle -(1+d)=-2(\pm\sqrt\frac{1}{2})$
$\displaystyle d=2(\pm\sqrt\frac{1}{2})-1$

N.B. When choosing values to input into the equation, if you let one of a,c,d be positive, then all the others have to be positive and vice-versa.
Possible values

3. Originally Posted by I-Think
It's the undertaker. Hi

a and c are the roots of the equation $\displaystyle x^2+ax-b=0$
Then $\displaystyle (x-a)(x-c)=x^2+ax-b$
$\displaystyle x^2-(a+c)x+ac=x^2+ax-b$
Equating coefficients
$\displaystyle -(a+c)=a$ $\displaystyle ac=-b....(1)$
$\displaystyle c=-2a.....(2)$

b and d are the roots of the equation $\displaystyle x^2+cx+d=0$
Then $\displaystyle (x-b)(x-d)=x^2+cx+d$
$\displaystyle x^2-(b+d)+bd=x^2+cx+d$
Equating coefficients
$\displaystyle -(b+d)=c...(3)$ $\displaystyle bd=d.....(4)$

We have a system of 4 equations
$\displaystyle ac=-b....(1)$ $\displaystyle c=-2a.....(2)$
$\displaystyle -(b+d)=c...(3)$ $\displaystyle bd=d.....(4)$

From (4):$\displaystyle bd=d$
$\displaystyle b=1$

Into (3):$\displaystyle -1-d=c$ and into (1): $\displaystyle ac=-1$
(2) into (1):$\displaystyle -2a^2=-1\longrightarrow a^2=1/2$
$\displaystyle a=\pm\sqrt\frac{1}{2}$

Hence c$\displaystyle =-2(\pm\sqrt\frac{1}{2})$

Into (4):$\displaystyle -(1+d)=-2(\pm\sqrt\frac{1}{2})$
$\displaystyle d=2(\pm\sqrt\frac{1}{2})-1$

N.B. When choosing values to input into the equation, if you let one of a,c,d be positive, then all the others have to be positive and vice-versa.
Possible values
a=b=c=d=0?

CB