Question

1)3x^2+2x=0 2)x^2-x+11=0 3)2x^2-18=0

Answer 1

1) 3x^2+2x=0

x*(3x+2)=0

x1=0

3x+2=0

x2= - 2/3

2) x^2-x+11=0

квадратное, решу через дискриминант

D=b^2-4ac= 1-4*11= -43

корней нет, тк D<0

3) 2x^2-18=0

2*(x^2-9)=0

x^2-3^2=0

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

x1= 3

x2= -3

Answer 2

To solve these quadratic equations, we can use the quadratic formula, which is given by:

For a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by: x = (-b ± √(b^2 - 4ac)) / 2a

Let's solve each of the equations:

1. 3x^2 + 2x = 0 Here, a = 3, b = 2, and c = 0 Using the quadratic formula: x = (-2 ± √(2^2 - 4 * 3 * 0)) / 2 * 3 x = (-2 ± √(4)) / 6 x = (-2 ± 2) / 6

The two solutions are: x = ( -2 + 2 ) / 6 = 0 x = ( -2 - 2 ) / 6 = -4/6 = -2/3

So the solutions to the equation are x = 0 and x = -2/3.

2. x^2 - x + 11 = 0 Here, a = 1, b = -1, and c = 11 Using the quadratic formula: x = (1 ± √((-1)^2 - 4 * 1 * 11)) / 2 * 1 x = (1 ± √(1 - 44)) / 2 x = (1 ± √(-43)) / 2

Since the discriminant (b^2 - 4ac) is negative, the solutions involve complex numbers.

The two solutions are: x = (1 + √43i) / 2 x = (1 - √43i) / 2

3. 2x^2 - 18 = 0 Here, a = 2, b = 0, and c = -18 Using the quadratic formula: x = (0 ± √(0^2 - 4 * 2 * -18)) / 2 * 2 x = (0 ± √(0 + 144)) / 4 x = (0 ± √144) / 4 x = (0 ± 12) / 4

The two solutions are: x = 12 / 4 = 3 x = -12 / 4 = -3

So the solutions to the equation are x = 3 and x = -3.

Note: For equation 2, since the discriminant is negative, the solutions involve complex numbers in the form of "a + bi", where "a" and "b" are real numbers, and "i" is the imaginary unit (√(-1)).