Question

Send it quick guys...

Answer 1

Answer:

194

Step-by-step explanation:

Recall 2 rules to solve this easily:

1. $(a+b)^2=a^2+2ab+b^2$

2. $(a-b)^2=a^2-2ab+b^2$

Now, if we let a = 7  and  $b=4\sqrt{3}$, we can say the problem is basically of the form:

$(a+b)^2 + (a-b)^2$

This can be simplified using the rules:

$(a+b)^2 + (a-b)^2\\a^2+2ab+b^2+a^2-2ab+b^2\\2a^2+2b^2$

Now we can plug the values of a and b we initially thought of (remember though  $\sqrt{x} \sqrt{x} =x$):

$2a^2+2b^2\\2(7)^2+2(4\sqrt{3})^2\\ 2(49)+2(4\sqrt{3})(4\sqrt{3})\\98+2(4)(4)(\sqrt{3} )(\sqrt{3} )\\98+2(4)(4)(3)\\98+96 \\194$

194 is the final answer.