P&C - 5

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x₁, x₂, x₃, x₄ are real nos.
x₁ < x₂ < x₃ < x₄.
Product of any three chosen at a time + the remaining = 130.

Find the sum of all possible values of x₄ rounded to the nearest decimal place.

#Mathematics #Algebra

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5 Answers

1133

Sourish Ghosh ·2013-03-22 08:26:15

correction: rounded to nearest integer? and relations should be â‰¤ and not <

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2305

Shaswata Roy ·2013-03-22 08:33:57

Is the answer 0?

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1133

Sourish Ghosh ·2013-03-22 08:39:45

No idea :P

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2305

Shaswata Roy ·2013-03-22 08:52:35

Let P=x₁x₂x₃x₄

x₁+x₂x₃x₄ = 130

x₁+Px₁=130

Similarly,

x₂+Px₂=130=x₁+Px₁

â†’x₁x₂ = P = x₁x₂x₃x₄

â†’x₃x₄=1

Similarly,

x₂x₃ = 1 = x₃x₄
â†’x₂ = x₄

Similarly we can find that:

x₁=x₂=x₃=x₄

Therefore,

x₁+x₁³=130

â†’x₁³+x₁-130=0

Using vieta's formula,

Sum of roots = -coeffcient of x²coeffcient of x³ = -01=0

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Anurag Roy I think it should be (x1)*(x4)=1;(x2)*(x3)=1;that satisfies the condition x1
Upvote·0· Reply ·2013-03-24 06:18:26
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1133

Sourish Ghosh ·2013-03-22 11:04:04

Ok the answer is supposed to be 395.

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