Evaluate sqrt of (2003)(2005)(2007)(2009)+16 by using algebraic method.
Let a = 2006. Then you have
(a-3)(a-1)(a+1)(a+3) = (a-1)(a+1)*(a-3)(a+3) = (a^2-1)(a^2-9)
Let b = a^2+4 = 2006^2+4. Now you have
= (b-5)(b+5) = b^2-25
Hmmm. Stuck here so far.
Anyone see a way out?
Oops. My bad.
Let b = a^2-5 = 2006^2-5. Now you have
= (b+4)(b-4) = b^2-16
So,
(b^2-16 + 16 = b^2 = (2006^2-5)^2
So,
√((2003)(2005)(2007)(2009)+16) = 2006^2-5 = 4024031
heroic. I was doing similar scratching for about half an hour :)
To evaluate √((2003)(2005)(2007)(2009) + 16) using an algebraic method, you can follow these steps:
Step 1: Notice that the given expression can be written as a difference of squares.
The expression can be simplified as: √(((2004)^2 − 1)(2004^2 + 15))
Step 2: Apply the algebraic identity a^2 - b^2 = (a + b)(a - b) to the expression.
Using the identity, we can rewrite the expression as: √((2004 + 1)(2004 - 1)(2004^2 + 15))
Step 3: Further simplify the expression.
Simplifying the expression, we have: √((2005)(2003)(2004^2 + 15))
Step 4: Notice that 2004^2 + 15 can be factored.
Factorizing 2004^2 + 15, we get: 2004^2 + 15 = (2004^2 + 30 + 15) - 30 = (2004 + 5)^2 - 30
Step 5: Substitute the factorized expression back into the original expression.
Substituting the factorized expression, we get: √((2005)(2003)((2004 + 5)^2 - 30))
Step 6: Simplify the expression further.
Simplifying the expression, we have: √((2005)(2003)((2009 - 30))
Step 7: Calculate the remaining factors.
Calculating the factors, we have: √(((4010)(6012))
Step 8: Calculate the square root.
Calculating the square root of the expression, we get: √(24,112,120)
Therefore, the final answer is: 4,911.28 (approximately).