Question

I just the answer of "ii"​

Answer 1

Answer:

1) (a ± b)2 = a2 ± 2ab + b2

3 - 2x - x2 =

3 - (2x + x2) =

3 - (x2 + 2 · 1 · x + 12 - 12) =

3 - [(x2 + 2x + 1) - 1] =

3 - [(x + 1)2 - 1] =

3 - (x + 1)2 + 1 =

4 - (x + 1)2

sorry I don't know answer for second

Answer 2

QUESTION:

TO FIND THE SQUARE OF

$1 \div \sqrt{2} ( \sqrt{3a} - \sqrt{5b} )$

SOLUTION:

$=( \sqrt{3a} - \sqrt{5b} ) \div \sqrt{2}$

$= ( \sqrt{6a} - \sqrt{10b} ) \div 2$

$on \: rationalising \: \sqrt{2} \: \: i.e. \: on \: \\ multiplying \: \sqrt{2} \: with \\ \: denominator \: and \: numerator \\ we \: get \: \\ \\ = ( \sqrt{6a} - \sqrt{10b} ) \div 2$

$so \: the \: final \: answer \: is \:$

$(\sqrt{6a} - \sqrt{10b} ) \div 2$