Question

If a^2+b^2+c^2=20 and a+b+c=8, find the value of ab+bc+ca?​

Answer 1

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{~~QUESTION~~}}}}}}}$

• If a^2+b^2+c^2=20 and a+b+c=8, find the value of ab+bc+ca?

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{!!ANSWER!!}}}}}}}$

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\large\mathcal{\bf{\boxed{\large\mathcal{~~~22~~}}}}}$

______________________________________________

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

$a {}^{2} + b {}^{2} + c {}^{2} = 20$

and

$a + b + c = 8$

now....the formula is ...

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

so therefore....

$ab + bc + ca = \frac{(a + b + c) {}^{2} - (a {}^{2} + b {}^{2} + c {}^{2} }{2} \\ = > ab + bc + ca = \frac{8 {}^{2} - 20}{2} \\ = > (ab + bc + ca) = 22$

$\huge\mathcal\green{\underline{hope\:\: this\:\: helps\:\: you}}$