Question

(x+3)(x-5)=-15 (3x-1)^2=(x-3)^2

Answer 1

Первое уравнения
$(x + 3)(x - 5) = - 15$
получиться если умножить
${x}^{2} + 7x = 0$
$x(x + 7) = 0$
x=0 или x = -7;
второе уравнения:
${(3x - 1)}^{2} = {(x - 3)}^{2}$
${(3x - 1)}^{2} - {(x - 3)}^{2} = 0$


$(3x - 1) - (x - 3) \times (3x - 1) + (x - 3)$
$2(x + 1) \times 4(x - 1) = 0$
${8(x - 1)}^{2} = 0$
$x - 1 = 0$
$x = 1$
икс равно один

Answer 2

Let's solve each equation separately:

1. (x+3)(x-5)=-15

Expanding the left-hand side of the equation:

x^2 - 2x - 15 = -15

Simplifying:

x^2 - 2x = 0

Factorizing:

x(x-2) = 0

Therefore, the solutions are:

x = 0 or x = 2

2. (3x-1)^2=(x-3)^2

Expanding the left-hand side and right-hand side of the equation:

9x^2 - 6x + 1 = x^2 - 6x + 9

Moving all terms to one side:

8x^2 - 8 = 0

Dividing by 8:

x^2 - 1 = 0

Factorizing:

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

Therefore, the solutions are:

x = 1 or x = -1

Therefore, the solutions to the system of equations are:

x = 0, x = 2, x = 1, or x = -1.