Question

if a+b+c=10,a2+b2+c2=29,find ab+bc+ca

Answer 1

Solution:

Given that

$a + b + c = 10 ...eq1\\ \\ {a}^{2} + {b}^{2} + {c}^{2} = 29 ...eq2$
To find
$ab + bc + ca$
Square eq1 both side

${(a + b + c)}^{2} = {(10)}^{2} \\ \\ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 100$
because we know that

${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz$
put the value of eq2

$29 + 2ab + 2bc + 2ca = 100 \\ \\ 2ab + 2bc + 2ca = 100 - 29 \\ \\ 2(ab + bc + ca) = 71 \\ \\ ab + bc + ca = \frac{71}{2} \\ \\ \\ ab + bc + ca = 35.5$

Answer 2

Answer:

Solution:

Given that

\begin{gathered}a + b + c = 10 ...eq1\\ \\ {a}^{2} + {b}^{2} + {c}^{2} = 29 ...eq2\\ \\\end{gathered}

a+b+c=10...eq1

a

2

+b

2

+c

2

=29...eq2

To find

\begin{gathered}ab + bc + ca \\ \\\end{gathered}

ab+bc+ca

Square eq1 both side

\begin{gathered}{(a + b + c)}^{2} = {(10)}^{2} \\ \\ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 100 \\ \\\end{gathered}

(a+b+c)

2

=(10)

2

a

2

+b

2

+c

2

+2ab+2bc+2ca=100

because we know that

\begin{gathered}{(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz \\ \\\end{gathered}

(x+y+z)

2

=x

2

+y

2

+z

2

+2xy+2yz+2xz

put the value of eq2

\begin{gathered}29 + 2ab + 2bc + 2ca = 100 \\ \\ 2ab + 2bc + 2ca = 100 - 29 \\ \\ 2(ab + bc + ca) = 71 \\ \\ ab + bc + ca = \frac{71}{2} \\ \\ \\ ab + bc + ca = 35.5 \\ \\\end{gathered}

29+2ab+2bc+2ca=100

2ab+2bc+2ca=100−29

2(ab+bc+ca)=71

ab+bc+ca=

2

71

ab+bc+ca=35.5