Question

(x^2-4)•√x-1=0 сумма корней уравнения P.s√x-1 под одним корнем

Answer 1

$(x^2-4)\sqrt{x-1}=0\; \; ,\; \; \; ODZ:\; x-1\geq 0\; ,\; x\geq 1\\\\(x-2)(x+2)\sqrt{x-1} =0\\\\x_2=2\; ,\; x_2=-2\; ,\; x_3=1\\\\x_1=2\geq 1\; ,\; \; \; \; x_2=-2$

Answer 2

To solve the equation (x^2 - 4) • √x - 1 = 0, we first need to find the values of x that satisfy the equation. Then, we'll find the sum of these roots.

Step 1: Solve the equation (x^2 - 4) • √x - 1 = 0

Let's consider each factor separately:

1. x^2 - 4 = 0 We can factor this as (x - 2)(x + 2) = 0 So, the solutions for this factor are x = 2 and x = -2.

2. √x - 1 = 0 Adding 1 to both sides, we get √x = 1 Squaring both sides, we get x = 1.

Step 2: Find the sum of the roots The roots we found in step 1 are x = 2, x = -2, and x = 1. Now, let's calculate the sum of these roots:

Sum of the roots = 2 + (-2) + 1 = 1

So, the sum of the roots of the equation is 1.