Question

Can someone help me with the 14•2 under radical sign. I don’t think it’s correct

Answer 1

`\bf (2√(7)+√(27))(√(28)-3√(3))\implies (2√(7)+√(9\cdot 3))(√(4\cdot 7)-3√(3)) \\\\\\ (2√(7)+√(3^2\cdot 3))(√(2^2\cdot 7)-3√(3))\implies (2√(7)+3√(3))(2√( 7)-3√(3))`

`\bf \stackrel{\underline{2√(7)* (2√( 7)-3√(3))}}{(2√(7))^2-6√(21)}+\stackrel{\underline{3√(3)*(2√( 7)-3√(3))}}{6√(21)-(3√(3))^2}\implies (2^2√(7^2))-6√(21)+6√(21)-(3^2√(3^2)) \\\\\\ (4\cdot 7)-0-(9\cdot 3)\implies 28-27\implies \boxed{1}`

notice, you can always multiply polynomials like say (a+b)(c+d+e), by simply doing a(c+d+e) + b(c+d+e), then combine like-terms and simplify.

Answer 2

Rationalize sqrt(28) by separating the two factors (4 and 7), so you get:

sqrt(4*7)

which becomes:

sqrt(4) * sqrt(7)

and you can simplify the square root of 4 to:

2*sqrt(7) ------ ANS