Question

if a+b+c=6 ab+bc+ac =11 find a^2+b^2+c^2

Answer 1

Answer:

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$

$= > {a}^{2} + {b}^{2} + {c}^{2} = {(a + b + c)}^{2} - 2(ab + bc + ca)$

Putting a+b+c=6 and ab+bc+ca=11,

$= > {a}^{2} + {b }^{2} + {c}^{2} = {6}^{2} - 2(11) = 36 - 22$

$= > {a}^{2} + {b}^{2} + {c}^{2} = 14$

Hope it helps.

Answer 2

Answer:

-4 Will be the answer I think so....