Question

1. -0,09+4х^2=0 2. (x+2)^2-49 (x-10)^2-25y^2

Answer 1

1)

$- 0.09 + 4 {x}^{2} = 0 \\ ({2x})^{2} - {0.3}^{2} = 0 \\ (2x - 0.3)(2x + 0.3) = 0 \\ x = ± \frac{3}{20} = ±0.15$

2)

${(x + 2)}^{2} - 49 = {(x + 2)}^{2} - {7}^{2} = (x + 2 - 7)(x + 2 + 7) = (x - 5)(x + 9)$

3)

${(x - 10)}^{2} - 25 {y}^{2} = {(x - 10)}^{2} - {(5y)}^{2} = (x - 10 - 5y)(x - 10 + 5y)$

Answer 2

Let's solve the equations one by one:

1. -0.09 + 4x^2 = 0

To solve this equation, we'll isolate the term with x^2 and then find the square root of both sides:

4x^2 = 0.09

Dividing both sides by 4:

x^2 = 0.09/4

x^2 = 0.0225

Taking the square root of both sides:

x = ±√0.0225

x ≈ ±0.15

Therefore, the solutions for this equation are x ≈ 0.15 and x ≈ -0.15.

2. (x + 2)^2 - 49 = (x - 10)^2 - 25y^2

Expanding both sides of the equation, we get:

x^2 + 4x + 4 - 49 = x^2 - 20x + 100 - 25y^2

Simplifying the equation:

4x - 45 = -20x - 25y^2

Moving all the x terms to one side and the y^2 terms to the other side:

4x + 20x = -25y^2 + 45

24x = -25y^2 + 45

Dividing both sides by 24:

x = (-25y^2 + 45)/24

So, the solution for this equation is x = (-25y^2 + 45)/24.