Question

412) 1,4x² - 68,6=0; 13) x² - 33 = 0; 14) 14x² + 140x = 0;

Answer 1

Ответ:

$12)~x_1=7, ~~x_2=-7;\\13) ~x_1=\sqrt{33},~~x_2=-\sqrt{33} ;\\14)~ x_1=-10,~~x_2=0$

Объяснение:

$12)~~ 1,4x^2-68,6=0;\\1,4(x^2-49)=0;\\x^2-49=0;\\x_{1,2}=\pm\sqrt{49} =\bf\pm7$

$13)~~x^2-33=0;\\x^2=33;\\x_{1,2}=\bf\pm\sqrt{33}$

$14)~~ 14x^2+140x=0;\\14x(x+10)=0;\\x_{1}=\bf-10;$$x_2=\bf0$

Answer 2

Ответ:

$1) \Large x_1 = 7,x_2 = -7$

$2) \Large x_1 = \sqrt{33} ,x_2 = - \sqrt{33}$

$3) \Large x_1 = 0,x_2 = - 10$

Объяснение:

$1) \large1.4x {}^{2} - 68,6 = 0$

$\large(1,4{x}^{2} - 68,6) : 1,4 = 0 :1,4$

$\large1,4 {x}^{2} : 1,4 - 68,6 : 1,4 = 0 \: \: \Rightarrow \: \: \: 1,4 {x}^{2} : 1,4 = \boxed{ {x}^{2} }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large - 68,6 : 1,4 = \boxed{ - 49}$

$\large {x}^{2} - 49 = 0$

$\large{x}^{2} = 49$

$\large x = \pm \sqrt{49}$

$\large x = \pm7$

$\Large \boxed{otvet :x_1 = 7,x_2 =- 7}$

$\large2) {x}^{2} - 33 = 0$

$\large {x}^{2} - 33 + 33 = 0 + 33 \: \: \: \Rightarrow \: \: \: - 33 + 33 = \boxed{0}\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large 0 + 33 = \boxed{33}$

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$\Large \boxed{otvet :x_1 = \sqrt{33} ,x_2 = - \sqrt{33} }$

$\large3)14 {x}^{2} + 140x = 0$

$\large14x(x + 10) = 0$

$\large x \cdot(x + 10) = 0$

$\Large \boxed{otvet : x_1 =-10,x_2 = 0}$

Answer 3

Let's solve each quadratic equation step by step:

1. 1. 4x² - 68.6 = 0

To solve this equation, we'll first move the constant term to the right side:

4x² = 68.6

Next, divide both sides of the equation by 4 to isolate x²:

x² = 68.6/4

x² = 17.15

Now, take the square root of both sides to solve for x:

x = ±√17.15

Therefore, the solutions to the equation 4x² - 68.6 = 0 are x = √17.15 and x = -√17.15.

2. x² - 33 = 0

To solve this equation, we'll move the constant term to the right side:

x² = 33

Taking the square root of both sides gives us:

x = ±√33

Therefore, the solutions to the equation x² - 33 = 0 are x = √33 and x = -√33.

3. 14x² + 140x = 0

We can factor out 14x from the equation:

14x(x + 10) = 0

Now, apply the zero product property, which states that if a product of factors is equal to zero, then at least one of the factors must be zero. Therefore, we have:

14x = 0 or x + 10 = 0

For the first equation, 14x = 0, divide both sides by 14:

x = 0

For the second equation, x + 10 = 0, subtract 10 from both sides:

x = -10

Therefore, the solutions to the equation 14x² + 140x = 0 are x = 0 and x = -10.