if a²+b²+c²=50 and ab+bc+ca=46, find a+b+c
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Step-by-step explanation:
We know that
[tex] {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)[/tex]
Given,
[tex] {a}^{2} + {b}^{2} + {c}^{2} = 50[/tex]
Also, ab+bc+ca=46
So,
[tex] {(a + b + c)}^{2} = 50 + 2(46) = 142[/tex]
So,
[tex]a + b + c = \sqrt{142} [/tex]
Thank you
[tex] \huge\red{\mathbb{ QUESTION}}[/tex]
★ ★ a²+b²+c²=50 and ab+bc+ca=46, find a+b+c
[tex] \huge\pink{ \mathbb{SOLUTION} \implies: } [/tex]
[tex] \huge \bf\ using \: formula \leadsto: \\ \\ \bf\ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ba)[/tex]
[tex] \sf\ {a}^{2} + {b}^{2} + {c}^{2} =50 \: and \: ab+bc+ca=46 \\ \\ \bf\ \implies: {(a + b + c)}^{2} =50 + 2(46) \\ \\ \bf\ \implies: {(a + b + c)}^{2} = 50 + 92 \\ \\ \bf\ \implies:{(a + b + c)}^{2} = 142 \\ \\ \bf\ \implies: a + b + c = \sqrt{142} \\ \\ \bf\ \implies:a + b + c = 11 \sqrt{21} [/tex]