Question

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

Answer 1

Answer:

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)