Question

1² + (√2-1)²
solve this ​

Answer 1

▪︎ GIVEN,

$\sf \mapsto {1}^{2} + ( \sqrt{2} - 1) {}^{2}$

▪︎ ID Used,

$\sf \mapsto(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

▪︎ Concept used,

$\sf \mapsto$ BODMAS.

(Bracket Open Division Multiplication Addition Subtracting)

▪︎ Now,

• By Applying BODMAS we have to first open the brackets.

$\sf \mapsto {1}^{2} - ( \sqrt{2} - 1) {}^{2} \\ \\ \sf = {1}^{2} - ( {( \sqrt{2}) }^{2} + (1) {}^{2} - 2( \sqrt{2})(1) \\ \\\sf = 1 {}^{2} - (2 + 1 - 2 \sqrt{2}) \\ \\ \sf = 1 {}^{2} - (3 - 2 \sqrt{2}) \\ \\ \sf = {1}^{2} - 3 + 2 \sqrt{2} \\ \\ \sf(by \: opening \: brackets) \\ \\ \sf = 1 - 3 + 2 \sqrt{2} \\ \\ \sf = - 2 + 2 \sqrt{2} \\ \\ \sf = 2 \sqrt{2} - 2 \\ \\ \sf\large\fbox\pink{ = 2( \sqrt{2} - 1)}$

◇ Therefore your answer is 2(root2 - 1).

Answer 2

Answer:

(2√2-5)²=33-20√2 by( a-b)²=a²-2ab+b²

(3√2+√3)²=21+6√6 by (a+b)²=a²+2ab+b²

(√2-1)²=3-2√2 by (a-b)²=a²-2ab+b²

=(33-20√2)+(21+6√6)-(3-2√2)

=33-20√2+21+6√6-3+2√2

=51-18√2+6√6

mark me