Let a,b,c be real and positive parameters. Solve the equation:
√(a+bx) + √(b+cx) + √(c+ax) = √(b-ax) + √(c-bx) + √(a-cx) Quick plz...
Answers
Answered by
7
x = 0 is the only solution.
LHS each term increases continuously (monotonously) as x increases.
RHS each term decreases as x increases. So RHS monotonously decreases as x increases.
Hence there can only be one intersection point of the two graphs of LHS and RHS. Only one solution.
See the diagram. LHS is like the graph of y = sqrt(x) but shifted left. RHS is the graph of y = sqrt (-x) shifted to right.
r = sqrt(a)+sqrt(b)+sqrt(c).
q = minimum of a/b, b/c, c/a.
p = minimum of b/a, c/b, a/c.
LHS each term increases continuously (monotonously) as x increases.
RHS each term decreases as x increases. So RHS monotonously decreases as x increases.
Hence there can only be one intersection point of the two graphs of LHS and RHS. Only one solution.
See the diagram. LHS is like the graph of y = sqrt(x) but shifted left. RHS is the graph of y = sqrt (-x) shifted to right.
r = sqrt(a)+sqrt(b)+sqrt(c).
q = minimum of a/b, b/c, c/a.
p = minimum of b/a, c/b, a/c.
Attachments:
kvnmurty:
:-)
Thx bt can you explain with an Attachment......
Would be helpful
ok. I will
Click on the red hearts thanks above
Hehe
Thx
why you asked me this question?
why you wanted some attachment
Coz........ I'm a beginner at functions and couldn't see the difference b/w increasing and decreasing functions just so easily..........
Answered by
0
Answer:
It is easy to see that x = 0 is a solution. Since the right hand side is a
decreasing function of x and the left hand side is an increasing function
of x, there is at most one solution.
Thus x = 0 is the only solution to the equation.
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