Real Solutions to $a-b+c-d=0; ab=cd; a^2-b^2+c^2-d^2=-24;a^2+b^2+c^2+d^2=50$

[anemone]

Find all real solutions to the system

$a-b+c-d=0$

$ab=cd$

$a^2-b^2+c^2-d^2=-24$

$a^2+b^2+c^2+d^2=50$

 

[kaliprasad]

We are given
a- b + c - d = 0 ... (1)

ab = cd ...(2)

$a^2 - b^2 + c^2 - d^2 = - 24$ ...(3)

$ a^2 + b^2 + c^2 + d^2 = 50 $...(4)

Spoiler

from (1)
a- b = d- c... (5)

square above and using (2)
$(a-b)^2 + 4ab = (c-d)^2 + 4cd$
or $(a+b)^2 = (c+d)^2$
so $a + b = c + d$ ... (6)

or $a+ b = -c - d$ ..(7)

from (5) and (6) a = d and b= c but it is not possible as LHS of (3) is zero which is contradiction

from (5) and (7) a = -c and b = - d

so we get from (3) and (4)

$a^2 - b^2 = - 12$

$a^2 + b^2 = 25$

add above to get $2 a^2 = 13$, subtract to get $2b^2 = 37$
this gives 4 set of solutions

(a,b,c,d) = $(\sqrt\frac{13}{2},\sqrt\frac{37}{2},-\sqrt\frac{13}{2},-\sqrt\frac{37}{2})$

or $(\sqrt\frac{13}{2},-\sqrt\frac{37}{2},-\sqrt\frac{13}{2},\sqrt\frac{37}{2})$

or $(-\sqrt\frac{13}{2},-\sqrt\frac{37}{2},\sqrt\frac{13}{2},\sqrt\frac{37}{2})$

or $(-\sqrt\frac{13}{2},\sqrt\frac{37}{2},\sqrt\frac{13}{2},-\sqrt\frac{37}{2})$

 

[anemone]

We are given
a- b + c - d = 0 ... (1)

ab = cd ...(2)

$a^2 - b^2 + c^2 - d^2 = - 24$ ...(3)

$ a^2 + b^2 + c^2 + d^2 = 50 $...(4)

Spoiler

from (1)
a- b = d- c... (5)

square above and using (2)
$(a-b)^2 + 4ab = (c-d)^2 + 4cd$
or $(a+b)^2 = (c+d)^2$
so $a + b = c + d$ ... (6)

or $a+ b = -c - d$ ..(7)

from (5) and (6) a = d and b= c but it is not possible as LHS of (3) is zero which is contradiction

from (5) and (7) a = -c and b = - d

so we get from (3) and (4)

$a^2 - b^2 = - 12$

$a^2 + b^2 = 25$

add above to get $2 a^2 = 13$, subtract to get $2b^2 = 37$
this gives 4 set of solutions

(a,b,c,d) = $(\sqrt\frac{13}{2},\sqrt\frac{37}{2},-\sqrt\frac{13}{2},-\sqrt\frac{37}{2})$

or $(\sqrt\frac{13}{2},-\sqrt\frac{37}{2},-\sqrt\frac{13}{2},\sqrt\frac{37}{2})$

or $(-\sqrt\frac{13}{2},-\sqrt\frac{37}{2},\sqrt\frac{13}{2},\sqrt\frac{37}{2})$

or $(-\sqrt\frac{13}{2},\sqrt\frac{37}{2},\sqrt\frac{13}{2},-\sqrt\frac{37}{2})$

Nice solution!(Cool)(Sun) I especially like how you observed if $(a+b)^2 = (c+d)^2$, then both $a+b=c+d$ or $a+b=-(c+d)$ hold true for general case but then we need to choose wisely which one only works for our case . I mean, we all know we should put the plus minus sign when taking square root, but it seems so brilliant to apply it in question such as this one, and I learned quite a bit from you, kali!

 

[kaliprasad]

Nice solution!(Cool)(Sun) I especially like how you observed if $(a+b)^2 = (c+d)^2$, then both $a+b=c+d$ or $a+b=-(c+d)$ hold true for general case but then we need to choose wisely which one only works for our case . I mean, we all know we should put the plus minus sign when taking square root, but it seems so brilliant to apply it in question such as this one, and I learned quite a bit from you, kali!

Thanks anemone, I checked for both cases and saw that one works and another does not. so I removed the erroneous case.

 

[CellCoree]

After analyzing the given system of equations, it can be observed that there are four variables (a, b, c, and d) and four equations. This suggests that there will be a unique solution for each variable.

To solve this system, we can use substitution and elimination methods. First, we can rearrange the first equation to get $a=b-d+c$. Then, we can substitute this value into the second equation to get $b(b-d+c)=cd$. Similarly, we can substitute the values of a and b into the third and fourth equations to get $a^2=(b-d+c)^2$, $b^2=b^2$, $c^2=(b-d+c)^2$, and $d^2=d^2$.

Next, we can use the equation $ab=cd$ to eliminate variables. Using the values obtained from substitution, we can rewrite this equation as $b(b-d+c)=(b-d+c)^2$. Simplifying this equation, we get $b^2-bd+bc=b^2-2bd+d^2+c^2-2cd+2bc$. Cancelling out the common terms, we get $bd=cd$. Dividing both sides by d, we get $b=c$.

Now, we can substitute the value of b=c into the first equation to get $a=c-d+c$. Simplifying this, we get $a=2c-d$. Similarly, we can substitute the values of a and b into the third and fourth equations to get $a^2=(2c-d)^2$, $b^2=c^2$, $c^2=(2c-d)^2$, and $d^2=d^2$.

Using the third and fourth equations, we can eliminate the variable d. Subtracting the third equation from the fourth equation, we get $c^2-(2c-d)^2=0$. Simplifying this, we get $c^2-4c^2+4cd-d^2=0$. Using the first equation, we can substitute $d=c-a+c$ into this equation to get $c^2-4c^2+4c(c-a+c)-(c-a+c)^2=0$. Simplifying this, we get $a^2-4ac+2c^2=0$.

Using the quadratic formula, we can solve for c. Substituting the values of a and