How do i algebraically solve |4x-3|=5√(x+4)

Step-by-step explanation:

|4x-3|=5√(x+4)  ⇔ |4x-3|²=5²(√(x+4))²  and x+4 ≥ 0

                         ⇔ (4x-3)² = 25(x+4)  and x+4 ≥ 0  ( because : /a/² = a²)

                         ⇔16x²-24x+9 = 25x +100  and x+4 ≥ 0

                         ⇔ 16x² -49x - 91 =0 and x+4 ≥ 0 quadratic equation

Δ = (-49)²-4(16)(-91) = 8225

two solution : X1 = (49-√8225)/32  ≅ - 1.3 accept (-1.3+4 ≥ 0)

                     X2 = (49+√8225)/32  ≅4.37 accept (4.37+4 ≥ 0)

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step-by-step explanation:

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: