Question

ifa+b+c=13 and a2 +b2+c2 = 69, find the value of ab + bc+ca.
a2 means a square ​

Answer 1

a+b+c = 13                  (equation 1)

a^2+b^2+c^2 = 69     (equation 2)

there is an identity that states :-

a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)

according to equation 1

a^2 + b^2 + c^2 = 69

so,  69 = (a + b + c)^2 - 2(ab + bc + ca)

according to equation 2

a+b+c = 13  

so,  69 = (13)^2 - 2(ab + bc + ca)

  ⇒ 69 = 169 - 2(ab + bc + ca)

  ⇒ 69 - 169 = -2(ab + bc + ca)

  ⇒ -100 = -2(ab + bc + ca)    

  ⇒ 2(ab + bc + ca) = 100  

  ⇒ 2(ab + bc + ca)/2 = 100/2           (dividing both the sides by 2)

  ⇒ ab + bc + ca = 50

Hope it Helps

Answer 2

Question :-

If a + b + c = 14 and a² + b² + c² = 69 . Find the value of ab + bc + ca ?

Answer :-

Given :-

a + b + c = 14

a² + b² + c² = 69

Required to find :-

• Value of a² + b² + c² ?

Identity used :-

$\huge{\dagger{\small{\boxed{\rm{ {(\;x + y + z )}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx }}}}}{\huge{\blue{\bigstar}}}$

Solution :-

Given that :-

a + b + c = 13 $\longrightarrow{\text{Equation 1 }}$

consider this as equation 1

a² + b² + c² = 26 $\longrightarrow{\text{Equation 2 }}$

we need to find the value of ab + bc + ca

So,

Let's consider equation 1

a + b + c = 13

squaring on both sides

( a + b + c )² = ( 13 )²

Now expand the L.H.S side according to identity

The identity is

( x + y + z )² = x² + y² + z² + 2xy + 2yz + 2xz

So,

a² + b² + c² + 2ab + 2bc + 2ca = 169

Now take 2 common on the left side

a² + b² + c² + 2 ( ab + bc + ca ) = 169

[ a² + b² + c² ] + 2 ( ab + bc + ca ) = 169

Now,

Substitute the value of a² + b² + c² from equation 2

So,

69 + 2 ( ab + bc + ca ) = 169

Transpose 69 to the right side

2 ( ab + bc + ca ) = 169 - 69

2 ( ab + bc + ca ) = 100

ab + bc + ca = 100/2

ab + bc + ca = 50

Hence,

Value of ( ab + bc + ca ) = 50

Points to remember :-

These questions can be solved by taking the given statements as hints.

Some of the most used identities are ;

• ( a + b )³ = a³ + b³ + 3a²b + 3ab²

• ( a - b )³ = a³ - b³ - 3a²b + 3ab²

• ( a + b )² = a² + b² + 2ab

• ( a - b )² = a² + b² - 2ab