Question
If $a^2+ b^2+ c^2= 35$ and $ab + bc + ca = 23;$ find $a+ b+ c.$
✓
Answer
We know that
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)\ldots(1)$
Given that, $a^2+b^2+c^2=35$ and $a b+b c+c a=23$
We need to find $a+b+c$ :
Substitute the values of $\left(a^2+b^2+c^2\right)$ and $(a b+b c+c a)$
in the identity $(1),$ we have
$(a+b+c)^2=35+2(23)$
$ \Rightarrow(a+b+c)^2=81$
$ \Rightarrow a+b+c= \pm \sqrt{81}$
$ \Rightarrow a+b+c$
$= \pm 9$
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)\ldots(1)$
Given that, $a^2+b^2+c^2=35$ and $a b+b c+c a=23$
We need to find $a+b+c$ :
Substitute the values of $\left(a^2+b^2+c^2\right)$ and $(a b+b c+c a)$
in the identity $(1),$ we have
$(a+b+c)^2=35+2(23)$
$ \Rightarrow(a+b+c)^2=81$
$ \Rightarrow a+b+c= \pm \sqrt{81}$
$ \Rightarrow a+b+c$
$= \pm 9$
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