Asked by Anish Panda
sqrt(8-3x)+sqrt(6x-1)=7-(2*(sqrt(9x-24)))
Answers
Answered by
Reiny
are we solving
√(8-3x) + √(6x-1) = 7 - 2√(9x-24) ????
I tested it in Wolfram, and there is no "real" solution, so I won't even try.
http://www.wolframalpha.com/input/?i=%E2%88%9A(8-3x)+%2B+%E2%88%9A(6x-1)+%3D+7+-+2%E2%88%9A(9x-24)
notice that the left side is only defined for
8-3x ≥ 0 AND 6x-1 ≥0
-3x ≥ -8 AND 6x ≥ 1
x ≤ 8/3 AND x ≥ 1/6
1/6 ≤ x ≤ 8/3 ----> the blue part of the graph
on the right side we have:
9x-24 ≥ 0
9x ≥ 24
x ≥ 8/3 , the only common value of the domains
if x = 8/3
RS = 7 - 2√0 = 7
LS = √0 + √15 ≠ 7
NO REAL SOLUTION!
√(8-3x) + √(6x-1) = 7 - 2√(9x-24) ????
I tested it in Wolfram, and there is no "real" solution, so I won't even try.
http://www.wolframalpha.com/input/?i=%E2%88%9A(8-3x)+%2B+%E2%88%9A(6x-1)+%3D+7+-+2%E2%88%9A(9x-24)
notice that the left side is only defined for
8-3x ≥ 0 AND 6x-1 ≥0
-3x ≥ -8 AND 6x ≥ 1
x ≤ 8/3 AND x ≥ 1/6
1/6 ≤ x ≤ 8/3 ----> the blue part of the graph
on the right side we have:
9x-24 ≥ 0
9x ≥ 24
x ≥ 8/3 , the only common value of the domains
if x = 8/3
RS = 7 - 2√0 = 7
LS = √0 + √15 ≠ 7
NO REAL SOLUTION!
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