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 $x^2+ax-b=0$
Then $(x-a)(x-c)=x^2+ax-b$
$x^2-(a+c)x+ac=x^2+ax-b$
Equating coefficients
$-(a+c)=a$ $ac=-b....(1)$
$c=-2a.....(2)$

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

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

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

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

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

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

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 $x^2+ax-b=0$
Then $(x-a)(x-c)=x^2+ax-b$
$x^2-(a+c)x+ac=x^2+ax-b$
Equating coefficients
$-(a+c)=a$ $ac=-b....(1)$
$c=-2a.....(2)$

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

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

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

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

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

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

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