Russia | A Nice Square Root Algebra Problem | Math Olympiad
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- Опубликовано: 8 сен 2024
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A very important formula in this case: a-b = (sqrt(a)-sqrt(b))*(sqrt(a)+sqrt(b)). a and b positive
Does this have any mathematical significance or use in any real situation? Not being critical, I would just really like to find out.
√[16(4+√15)]=√[64+16√15]
a²+2ab+b² a²+b²=64 2ab=16√15 ab=8√15
8√15=√960=√[2*2*2*2*2*2*3*5] a²=40 b²=24 a=40=2√10 b=24=2√6
√[a²+2ab+b²]=√[a+b]²=a+b
Ans : 2√10+2√6=2√2(√5+√3)
Nicely symmetric result: 4√(4+√(4+1)*√(4-1)) [multiply numerator and denominator by 4+√(15), and simplify...]
i have a even easily way to solve
√(16/4-√5)
= √(4-√5)
= √4 - √5
= 2 - √5
Therefore, √(16/4-√5) = 2 - √5.
This is clearly wrong and it has nothing to do with the problem in the clip.
√(16/4-√5) = √(4-√5) = 1.328131 > 0, but 2 - √5 =-0.236068
What is the point?. What is the difference between sqrt(16/(4-sqrt(15)) and 2*sqrt(2)*(sqrt(5)+sqrt(3)). I don't see that the expression is somewhat simplified.
The problem is to find square root of 16/(4-sqrt(15))
Looks not much simpler than at the beginning..
Olympiad Math: √[16/(4 - √15) = ?
16/(4 - √15) = 16(4 + √15) = 8(8 + 2√15) = 8[(√5)² + 2(√5)(√3) + (√3)²]
= 8(√5 + √3)²; √[16/(4 - √15) = √[8(√5 + √3)²] = 2√10 + 2√6
Answer check:
√[16/(4 - √15) = 2√10 + 2√6; Confirmed as shown
Final answer:
Simplified form; √[16/(4 - √15) = 2√10 + 2√6
asnwer=3/2 isit
asnwer=2/2+3/5 isit
Looks not much simpler than at the beginning...
I agree