Question

please solve this
I will mark you as BRAINLIEST​

Answer 1

Please mark me in the brainlist ✌️✌️✌️✌️✌️✌️

Please see the attachment below

Answer 2

AnswEr:-

• 41. (Que 6)

• 58. (Que 7).

ExplanaTion:-

Given:-

• 1. (a-b) = 5 and ab = 16.

• 2. (a+b+c) = 9 and ab+bc+ac = 23.

To Find:-

• 1. a² + b².

• 2. a² + b² + c².

IDs Used:-

$\\ \tt\longrightarrow (x +y) {}^{2} = {x}^{2} + {y}^{2} - 2xy. \\ \\ \tt\longrightarrow( x - y) ^{2} + 2xy = ( {x}^{2} + {y}^{2} - 2xy) + 2xy. \\ \\ \tt\longrightarrow(x - y) {}^{2} + 2xy = {x}^{2} \: + \: {y}^{2} \: \cancel{ - 2xy} \: \cancel{ + 2xy}. \\ \\ \hookrightarrow \boxed{ \boxed{ \bf (x - y) {}^{2} + 2xy = {x}^{2} + {y}^{2} . }}_{... \bf \: id(1).}$

And,

$\\ \tt\longrightarrow(x + y + z) {}^{2} = {x}^{2} + {y}^{2} + z {}^{2} + xy + yz + zx. \\ \\ \tt\longrightarrow(x + y + z) {}^{2} - (xy + yz + zx) = {x}^{2} + {y}^{2} + z {}^{2} \: \cancel {+( xy + yz + zx)} \: \cancel{ - (xy + yz + zx)}. \\ \\ \hookrightarrow \boxed{ \boxed{ \bf(x + y + z) {}^{2} - (xy + yz + zx) = {x}^{2} + {y}^{2} + {z}^{2}.}}_{... \bf \: id(2).}$

1)

So here,

• x = a.

• y = b.

By putting the values in id(1) we get:-

$\\ \tt\mapsto(x - y) {}^{2} + 2xy = {x}^{2} + {y}^{2} . \\ \\ \tt\mapsto {a}^{2} + {b}^{2} = (a - b) {}^{2} + 2ab. \\ \\ \tt\mapsto {a}^{2} + {b}^{2} = (5) {}^{2} + 2(16). \\ \\ \tt\mapsto {a}^{2} + {b}^{2} = 25 + 32. \\ \\ \bf\mapsto {a}^{2} + {b}^{2} = 57.$

Therefore the required value of a²+b² = 57

2)

So Here,

• x = a.

• y = b.

• z = c.

By putting the values in id(2) we get:-

$\\ \tt\mapsto {x}^{2} + {y }^{2} + z {}^{2} = (x + y + z) {}^{2} - (xy + yz + zx). \\ \\ \tt\mapsto {a}^{2} + {b}^{2} + c {}^{2} = (9) {}^{2} - (23). \\ \\ \tt\mapsto {a}^{2} + {b}^{2} + c {}^{2} = 81 - 23. \\ \\ \bf\mapsto {a}^{2} + {a}^{2} + c {}^{2} = 58.$

Terefore the value of a²+b²+c² = 58.