SOLUTION: find the real solutions of the equation. x^(2/5)-3x^(1/5)-4=0
Algebra.Com
Question 948047: find the real solutions of the equation. x^(2/5)-3x^(1/5)-4=0
Found 2 solutions by CubeyThePenguin, greenestamps:
Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
Let x^(1/5) = b.
y^2 - 3y - 4 = 0
(y-4)(y+1) = 0
y = 4, y = -1
x^(1/5) = 4, x^(1/5) = -1
The "-1" solution is extraneous, so x = 4^5 = 1024.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
And yet another hasty, careless, and incorrect response from tutor @CubeyThePenguin.
You can introduce a dummy variable, as he does in his response, to help solve the problem if you want; it is not necessary.
or
or
The x = -1 solution is NOT extraneous; it satisfies the original equation.
RELATED QUESTIONS
Find the number of real -number solutions of each equation
1. x^2+8=0
2. 3x^2-9x=-5
(answered by lynnlo)
1/x-1 + 1/x+2 = 5/4
Find all real solutions of the... (answered by Nate)
Find all real solutions of the equation
(x)square/x-1 + 1/x+2 = 5/4 (answered by Nate)
Find all real solutions to the equation 4^x - 5(2^x) - 24 =... (answered by Alan3354)
x^2+5=0
Determinbe the nature of the solutions of the equation. Choose from
2 real... (answered by ewatrrr)
Find the real solutions of each equation.
Let rt = root
1. x^2 - 3x - rt{x^2 - (answered by ikleyn)
Find all the real solutions of the equation.... (answered by jsmallt9)
find the real solutions, using the quadratic equation... (answered by vleith)
determine nature of solutions of the equation.
x^2+3x+5=0
a) two imaginary-number... (answered by josgarithmetic)