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)

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph