Question

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

Answer 1

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)