Cassarrim1

2022-02-01

Determine the real number A such that the following equation has a real answer for C for every real $B:{A}^{2}+{B}^{2}+{C}^{2}+2ABC=1$

sphwngzt

Expert

Step 1
Solve for C:
$C=-AB±\sqrt{{A}^{2}{B}^{2}-{A}^{2}-{B}^{2}+1}$
We need a real A such that
${A}^{2}{B}^{2}-{A}^{2}-{B}^{2}+1\ge 0$
for any real B. Notice that we can factor this to get
$\left({A}^{2}-1\right)\left({B}^{2}-1\right)\ge 0.$
Since
${B}^{2}-1\ge 0$ for all $|B|\ge 1$ and ${B}^{2}-1<0$ for all $|B|<1$ there are only two A-values that can maintain the above inequality regardless of the B-value; that is,
$A=±1$

Souticexi

Expert

Step 1
Whichever variable we solve for, we discover the same thing.
${A}^{2}+{B}^{2}+{C}^{2}+2ABC=1$
${A}^{2}+\left(2BC\right)A+\left({B}^{2}+{C}^{2}-1\right)=0$
$A=\frac{-2BC±\sqrt{{\left(2BC\right)}^{2}-4\left(1\right)\left({B}^{2}+{C}^{2}-1\right)}}{2}$
$A=-BC±\sqrt{\left({B}^{2}-1\right)\left({C}^{2}-1\right)}$
Here, we can see that both variables under that radical must be greater than one of less than one. Since the variables are interchangeable, we know that all three must share this property so, either

Do you have a similar question?