SOLUTION: I need detail solution: A = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} + {{{sqrt(5)}}})( {{{sqrt(2)}}} &#8722; {{{sqrt(3)}}} + {{{sqrt(5)}}}) and B = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} &#8722;

Click here to see ALL problems on Evaluation Word Problems

Question 824392: I need detail solution:
A = ( + + )( − + ) and B = ( + − )(− + + ) find C=AB
(In the "A" I've got 4-2, but I can't solve B)
Found 3 solutions by rothauserc, MathTherapy, math_tutor2020:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website!
A) let u = square root (2) + square root (5)
then we can rewrite problem as
(u + square root(3)) * (u - square root(3))
u^2 - 3
(2 + 2*square root(2)*square root(5) +5) - 3
(2 +2*square root(10) +5 -3)
(4 +2*square root(10))
2*(2 + square root(10))
B) let v = square root(2) - square root (5)
then we can rewrite problem as
(square root(3) + v) * (square root(3) - v)
3 - v^2
3 - (2 -2*square root(2)*square root(5)+5)
3 - (2 -2*square root(10)+5)
3 - 2 +2*square root(10) - 5
2*square root(10) - 4
2*(square root(10) - 2)
now we are asked what is C = AB
C = (2*square root(10) + 4) * (2*square root(10) -4)
C = 4*10 - 16
C = 40 -16
C = 24

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

I need detail solution: A = ( + + )( − + ) and B = ( + − )(− + + ) find C=AB (In the "A" I've got 4-2, but I can't solve B) ************************************************** A = B = C = OR A = ( + + )( - + ) A = A = + + A = + + A = ----- COLLECTING like-terms A = = <=== CORRECT value of A, so yours is INCORRECT B = A = B = C = AB C = C = ----- FOILing right-side C = - 16 + 4(10) = - 16 + 40 = 24

Answer by math_tutor2020(3838)   (Show Source):
You can put this solution on YOUR website!

Let and so they can be used in a substitution step.

This means

condenses temporarily to

Meanwhile

shortens to

or

after rearranging a few negative signs.

Then,

Use the difference of squares rule.

For a moment I'll put this main computation on pause.
Let's take a slight detour.

I'll rewrite one of the factors shown above.
I'll start with D and work my way up to D^2-5.

Through very similar scratch work, you should have become

Now returning to the main event.

I used GeoGebra to verify the answer is correct.
Another tool you could use is WolframAlpha.

SOLUTION: I need detail solution: A = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} + {{{sqrt(5)}}})( {{{sqrt(2)}}} &#8722; {{{sqrt(3)}}} + {{{sqrt(5)}}}) and B = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} &#8722;

SOLUTION: I need detail solution: A = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} + {{{sqrt(5)}}})( {{{sqrt(2)}}} − {{{sqrt(3)}}} + {{{sqrt(5)}}}) and B = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} −