SOLUTION: I have been trying for several hours to solve the following:
(20+sqrt5-x) = (sqrtx+10)
I have come to an answer of x^2-390x+154025 and x^2+5x+37006.25 but when I have tried t
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-> SOLUTION: I have been trying for several hours to solve the following:
(20+sqrt5-x) = (sqrtx+10)
I have come to an answer of x^2-390x+154025 and x^2+5x+37006.25 but when I have tried t
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Question 21436: I have been trying for several hours to solve the following:
(20+sqrt5-x) = (sqrtx+10)
I have come to an answer of x^2-390x+154025 and x^2+5x+37006.25 but when I have tried to factor both of these, I haven't been able to so I can find out what x is equal to.
Can you help? Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! (20+sqrt5-x) = (sqrtx+10)...I ASSUME THE PROBLEM IS
20 +(5-X)^0.5=(X+10)^0.5...IF THIS IS SO IT HAS NO SOLUTION AND DONT BREAK YOUR HEAD ON THAT.BECAUSE X CANNOT BE MORE THAN FIVE OR LESS THAN -10,AS THEN WE GET IMAGINARY NUMBERS AND IN THAT RANGE OF X BETWEEN -10 AND +5 THERE IS NO SOLUTION WHICH SATISFIES THAT EQN.AT X=5 WE GET LHS =20..AND RHS CANNOT BECOME 20..AT X=-10 RHS IS ZERO AND LHS CANNOT BE ZERO
IF NOT PLEASE WRITE THE PROBLEM CORRECTLY BY PUTTING APPROPRIATE BRACKETS...FOR.EX.WHETHER YOU MEAN
(sqrtx+10)=10+SQRT(X)...OR...SQRT(X+10)..SIMILARLY
(20+sqrt5-x)=20-5+SQRT(X)...OR....20+SQRT(X-5)
THEN I SHALL WORK OUT THE PROBLEM
