# Find all a,b,c \in \mathbb{R} that satisfy both equations: a+b+c=63 ab+bc+ac=2021

Find all a,b,c $\in \mathbb{R}$ that satisfy both equations:
a+b+c=63
ab+bc+ac=2021
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Step 1
The given equations are a+b+c=61 and ab+bc+ac=2021.
The objective is to find the real values of a, b, c that satisfy the above equations.
Step 2
The Square of a Trinomial formula stated as follows.
${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$
Rewrite the formula as follows.
${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2\left(ab+bc+ca\right)$
${\left(a+b+c\right)}^{2}-2\left(ab+bc+ca\right)={a}^{2}+{b}^{2}+{c}^{2}$
${a}^{2}+{b}^{2}+{c}^{2}={\left(a+b+c\right)}^{2}-2\left(ab+bc+ca\right)$
Substitute a+b+c=63 and ab+bc+ac=2021 in the above formula to find the unknowns if exists.
${a}^{2}+{b}^{2}+{c}^{2}={\left(63\right)}^{2}-2\left(2021\right)$
=3969-4042
${a}^{2}+{b}^{2}+{c}^{2}=-73$
It is known that, the square of any real number value is positive and obviously the sum of its squares also positive.
Note that, ${a}^{2}+{b}^{2}+{c}^{2}=-73$. That is, the sum of squares of the required numbers are negative.
Therefore, there is no such real number exist that satisfy the equations a+b+c=63 and ab+bc+ac=2021.