Question

Q.36 – If a+b= 16, and ab=10 find.
i. a2– 6ab + b 2​

Answer 1

Answer:

Q.36 – If a+b= 16, and ab=10 find.

i. a2– 6ab + b 2

Step-by-step explanation:

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Answer 2

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

$puttng \: the \: value \: of \: (a + b) = 16\: and \: ab = 10$

$=> \: {16}^{2} = {a}^{2} + {b}^{2} + 2(10)$

$=> \: 256 = {a}^{2} + {b}^{2} + 20$

$=> \: {a}^{2} + {b}^{2} = 256 - 20$

$=> \: {a}^{2} + {b}^{2} = 236$

Now according to question:—

${a}^{2} - 6ab + {b}^{2}$

$= {a}^{2} + {b}^{2} - 6ab$

$= \: 236 - 6(10)$

$= 236 - 60$

$= 176$

Hope it helps you !!