Question

X²-1-12=0 x²-7x+12=0

Answer 1

1).
${x}^{2} - 1 - 12 = 0$
${x}^{2} - 13 = 0$
${x }^{2} = 13$
$x = + - \sqrt{13}$
ответ : х1 = √13; х2 = -√13.

2).
${x }^{2} - 7x + 12 = 0$
$d = {b}^{2} - 4ac = 49 - 4 \times 1 \times 12 = 1$
$x1 = \frac{ - b + \sqrt{d} }{2a} = \frac{7 + 1}{2} = 4$
$x2 = \frac{ - b - \sqrt{d} }{2a} = \frac{7 - 1}{2} = 3$

Answer 2

Я так понимаю в 1

х^2-1х-12=0

Д=1-4(-12)=1+48=49

Х1=(1+7)/2=4

Х2=(1-7)/2=-3

Ну а если х^2-1-12=0,то

х^2-13=0

х^2=13

Значит Х равен +- корень из 13

х^2-7х+12=0

Д=49-4(12)=49-48=1

Х1=(7+1)/2=4

Х2=(7-1)/2=3

Answer 3

To solve the equations you provided, let's solve them one by one:

Equation 1: x² - 1 - 12 = 0

Combining like terms, we have: x² - 13 = 0

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 1, b = 0, and c = -13.

Substituting these values into the formula, we get: x = (0 ± √(0² - 4(1)(-13))) / (2(1)) x = (0 ± √(0 + 52)) / 2 x = (0 ± √52) / 2 x = ±√52 / 2 x = ±2√13 / 2 x = ±√13

Therefore, the solutions for Equation 1 are x = √13 and x = -√13.

Equation 2: x² - 7x + 12 = 0

This quadratic equation can be factored as: (x - 3)(x - 4) = 0

Setting each factor to zero and solving, we have: x - 3 = 0 => x = 3 x - 4 = 0 => x = 4

Therefore, the solutions for Equation 2 are x = 3 and x = 4.