Question

If a + b + c = 9 and ab + bc + ac = 40, then find the value for : a^2 + b^2 + c^2​

Answer 1

Given:-

☆Value of a+b+c=9

☆Value of ab+bc+ac=40

To find:-

☆Value of a²+b²+c²

Solution:-

By squaring both sides,in the equation a+b+c=9,we get:-

=>(a+b+c)²=(9)²

We know that:-

=>(a+b+c)²=a²+b²+c²+2(ab+bc+ac)

=>a²+b²+c²+2(ab+bc+ac)=81

=>a²+b²+c²+2(40)=81

=>a²+b²+c²+80=81

=>a²+b²+c²=81-80

=>a²+b²+c²=1

Thus,value of a²+b²+c² is 1.

Some extra information:-

The common identities used to solve this type of problems are:-

•(a+b)²=a²+b²+2ab

•(a-b)²=a²+b²-2ab

•a²-b²= (a-b)(a+b)

•(a+b)³=a³+b³+3ab(a+b)

•(a-b)³=a³-b³-3ab(a-b)

•a³+b³=(a+b)(a²+b²-ab)

•a³-b³=(a-b)(a²+b²+ab)

•(a+b+c)²=a²+b²+c²+2(ab+bc+ac)

Answer 2

Solution :

Given that,

=> a + b + c = 9

on squareing both side we get

=> a² + b² + c² = 9²

=> a² + b ² + c² + 2ab + 2bc + 2ac = 81

=> a² + b² + c² + 2(ab + bc + ac) = 81

on putting ab + bc + ac = 40 we get

=> a² + b² + c² + 2 x 40 = 81

=> a² + b² + c² = 81 - 80

=> 1

=> value for a² + b² + c² is 1