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alg

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i need help solving this one (x + 4)^2/3= 9 a) 13 b) 23;-31 c) 5 d) 31;-31

also sove check for extraneous solutions
(6x-5)^1/2 = (7+ 4x)^1/2
a) 1/6 b) -9 c) 6 or d) 1/5

  • alg -

    (x + 4)^2/3= 9
    [(x + 4)^2/3]^(3/2)= 9^(3/2)

    x+4 = 27
    x = 23

    do the second one by squaring both sides

  • alg -

    ok so for this one i need help solving this one (x + 4)^2/3= 9 a) 13 b) 23;-31 c) 5 d) 31;-31
    it would be c 23 or -31 right?
    for for the 2nd one i got -9?
    but i don't think i did it right

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