Question

1) -x=√x, 2) x²-2x-3=0, 3) 10x²=7x-1

Answer 1

Объяснение:

1. решений нет или же 0

2. D=b²-4ac=2²-4*1*(-3)=4+12=16=4²

x1=2-4/2=-1

x2=2+4/2=3

3.10x²-7x+1=0

D=b²-4ac=7²-4*10*1=49-40=9=3²

x1=7-9/20=-0,1

x2=7+9/20=4/5

Answer 2

Ответ:

1)

x = -$\sqrt{x}$

$x^{2}$+x = 0

x(x+1) = 0

x = 0 или х+1 = 0

                х = -1.

х = 0; 1

2.

$>-3x+x-3 = 0</p> <p>x(x-3)+(x-3) = 0</p> <p>(x+1)(x-3) = 0</p> <p>x+1 = 0          x-3 = 0</p> <p>x = -1             x = 3</p> <p>x = -1; 3</p> <p></p> <p>3.</p> <p>10<img src=$-7x+1 = 0

10$x^{2}$-2x-5x+1 = 0

2x(5x-1)-(5x-1) = 0

(5x-1)(2x-1) = 0

5x-1 = 0         2x-1 = 0

5x = 1            2x = 1

x = 1/5           x = 1/2

Answer 3

To solve these equations, let's go through them one by one:

1. -x = √x:

This equation doesn't have any real solutions. It's clear that x cannot be negative because the left side is negative, and the right side is the square root of x, which is always non-negative.

2. x² - 2x - 3 = 0:

You can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = -2, and c = -3.

x = (2 ± √((-2)² - 4(1)(-3))) / (2(1)) x = (2 ± √(4 + 12)) / 2 x = (2 ± √16) / 2 x = (2 ± 4) / 2

Now, we have two possible solutions:

x₁ = (2 + 4) / 2 = 6 / 2 = 3 x₂ = (2 - 4) / 2 = -2 / 2 = -1

So, the solutions for the equation x² - 2x - 3 = 0 are x = 3 and x = -1.

3. 10x² = 7x - 1:

First, we need to move all terms to one side of the equation to set it equal to zero:

10x² - 7x + 1 = 0

Now, you can solve this quadratic equation. You can use the quadratic formula again:

a = 10, b = -7, and c = 1.

x = (-b ± √(b² - 4ac)) / (2a)

x = (7 ± √((-7)² - 4(10)(1))) / (2(10)) x = (7 ± √(49 - 40)) / 20 x = (7 ± √9) / 20 x = (7 ± 3) / 20

Now, we have two possible solutions:

x₁ = (7 + 3) / 20 = 10 / 20 = 1/2 x₂ = (7 - 3) / 20 = 4 / 20 = 1/5

So, the solutions for the equation 10x² = 7x - 1 are x = 1/2 and x = 1/5.