Question

Решите Уравнения: а)4х²-9=0 б) 2y+y²=0

Answer 1

a) 4x² - 9 = 0

(2x - 3)(2x + 3) = 0

или   2x - 3 = 0              или   2x + 3 = 0

        2x = 3                              2x = - 3

        x₁ = 1,5                             x₂ = - 1,5

Ответ : - 1,5 ; 1,5

б)    2y + y² = 0

       y(2 + y) = 0

или   y₁ = 0          или   2 + y = 0

                                     y₂ = - 2

Ответ : - 2 ; 0

Answer 2

Ответ:

а)1,5 и б) -2; 0

Объяснение:

а) 4х²-9=0

4х² =9

х² = 9/4

x = 3/2 = 1,5

Ответ: 1,5

б) 2y+y²=0

y( 2+y) =0

y= 0   ; 2+ y =0

           y= -2

Ответ: -2; 0

Answer 3

Sure thing! Let's solve these equations:

а) $4x^2 - 9 = 0$

We can factor this equation as a difference of squares:

$(2x)^2 - 3^2 = 0$

$(2x + 3)(2x - 3) = 0$

This gives us two possible solutions:

$2x + 3 = 0$ or $2x - 3 = 0$

If we solve these, we get:

$2x = -3$ or $2x = 3$

So, $x = -\frac{3}{2}$ or $x = \frac{3}{2}$

б) $2y + y^2 = 0$

This equation can be factored as well:

$y(2 + y) = 0$

This gives us two possible solutions:

$y = 0$ or $2 + y = 0$

If we solve the second one, we get:

$y = -2$

So, the solutions are $y = 0$ or $y = -2$.