Question

Решить уравнение способом выделения квадрата двучлена 1. x^2+2x-3=0 2. x^2-6x-7=0 Решите какое сможете

Answer 1

1) ( x + 3)(x - 1)= 0
Произведение равно 0,когда один из множителей равен 0,значит,
x + 3 = 0
x= - 3
x - 1 = 0
x = 1
2) ( x - 7)(x + 1) = 0
x - 7 = 0
x = 7
x + 1 =0
x = - 1

Answer 2

Solving the Equations Using Completing the Square Method

Let's solve the given equations using the method of completing the square.

1. Equation: x^2 + 2x - 3 = 0

To solve this equation, we can use the completing the square method. First, we add 3 to both sides to get the constant term on the right side:

x^2 + 2x = 3

Now, we complete the square by adding the square of half the coefficient of x to both sides:

x^2 + 2x + (2/2)^2 = 3 + (2/2)^2 x^2 + 2x + 1 = 3 + 1 (x + 1)^2 = 4

Taking the square root of both sides gives us:

x + 1 = ±2 x = -1 ± 2 x = -3 or x = 1

So, the solutions for the equation x^2 + 2x - 3 = 0 are x = -3 and x = 1.

2. Equation: x^2 - 6x - 7 = 0

Similarly, for this equation, we add 7 to both sides to get the constant term on the right side:

x^2 - 6x = 7

Completing the square by adding the square of half the coefficient of x to both sides:

x^2 - 6x + (-6/2)^2 = 7 + (-6/2)^2 x^2 - 6x + 9 = 7 + 9 (x - 3)^2 = 16

Taking the square root of both sides gives us:

x - 3 = ±4 x = 3 ± 4 x = -1 or x = 7

So, the solutions for the equation x^2 - 6x - 7 = 0 are x = -1 and x = 7. These are the solutions for the given equations using the completing the square method.