Question

simplify the expression (6-√ 11)(6+√ 11)

Answer 1

Just multiply the two bracketted terms together. This will remove the radical signs and simplify.

Answer 2

What didyou get when you multiplied the bracketted terms?

Answer 3

To multiply using the FOIL method:
\[F\ pair=6\times 6=36\]
\[O\ pair=6\times \sqrt{11}=6\sqrt{11}\]
\[I\ pair=6\times -\sqrt{11}=-6\sqrt{11}\]
\[L\ pair=-\sqrt{11}\times \sqrt{11}=-11\]
When you add these results you get
\[36+6\sqrt{11}-6\sqrt{11}-11=you\ can\ calculate\]

Answer 4

Alternative Way: Use the difference of squares rule


\[\Large (a-b)(a+b) = a^2 - b^2\]


\[\Large (6-\sqrt{11})(6+\sqrt{11}) = 6^2 - \left(\sqrt{11}\right)^2\]


\[\Large (6-\sqrt{11})(6+\sqrt{11}) = 36 - 11\]


\[\Large (6-\sqrt{11})(6+\sqrt{11}) = 25\]

Answer 5

both lead to the same answer, just two different ways to get there

Answer 6

is it

\[\Large w = \sqrt{7w}\]

or is it

\[\Large w = \sqrt{7}w\]

the difference between the two is the placement of w

Answer 7

is the second w in the square root?

Answer 8

ok you would square both sides to get

\[\Large w^2 = 7w\]

what's next?

Answer 9

\[\Large w^2 = 7w\]

\[\Large w^2 - 7w = 0\]

\[\Large w(w - 7) = 0\]

how about now?

Answer 10

well let's check

Answer 11

plug in w = 0 (first possible solution)

\[\Large w = \sqrt{7w}\]

\[\Large 0 = \sqrt{7*0}\]

\[\Large 0 = \sqrt{0}\]

\[\Large 0 = 0\]

Since that last equation is true, w = 0 is indeed a solution

Answer 12

I'll let you check w = 7

Answer 13

you are correct

plug in w = 7 (second possible solution)

\[\Large w = \sqrt{7w}\]

\[\Large 7 = \sqrt{7*7}\]

\[\Large 7 = \sqrt{49}\]

\[\Large 7 = 7\]

that's true, so w = 7 is also a true solution

Answer 14

you're welcome