Question

1)4^2-20=0 2)3x^2+5x=0 3) x^2-5x-24=0 4)2x^2+13x=6=0 5)7x^2-6x+2=0 6)4x^2+12x+9=0

Answer 1

1) 4²-20=0

16-20=0

-4 =/= 0

2) 3x²+5x=0

x(3x+5)=0

x=0    3x+5=0

x=0    x=$-\frac{5}{3}$

3) x²-5x-24=0

x²+3x-8x-24=0

x(x+3)-8(x+3)=0

(x+3)(x-8)=0

x+3=0   x-8=0

x=-3      x=8

4)2x²+13x+6=0

2x²+12x+x+6=0

2x(х+6)+(х+6)=0

(x+6)(2x+1)=0

x+6=0   2x+1=0

x=-6      2x=-1

            x=-$\frac{1}{2}$

5) 7x²-6x+2=0

(тут какая-то ошибка)

6)4x²+12x+9=0

(2x+3)²=0

2x+3=0

2x=-3

x=-3:2

x=-1,5

Answer 2

To solve each of these quadratic equations, we can use the quadratic formula:

For an equation in the form of ax^2 + bx + c = 0, the quadratic formula is:

x = (-b ± √(b² - 4ac)) / (2a)

Let's solve each of the equations you provided:

1. 4^2 - 20 = 0

This equation doesn't appear to be quadratic. It simplifies to 16 - 20 = -4, which is not equal to zero.

2. 3x^2 + 5x = 0

a = 3, b = 5, and c = 0 in this equation.

Using the quadratic formula:

x = (-5 ± √(5² - 4 * 3 * 0)) / (2 * 3) x = (-5 ± √25) / 6 x = (-5 ± 5) / 6

There are two solutions:

x₁ = (-5 + 5) / 6 = 0/6 = 0 x₂ = (-5 - 5) / 6 = -10/6 = -5/3

So, the solutions are x = 0 and x = -5/3.

3. x^2 - 5x - 24 = 0

a = 1, b = -5, and c = -24 in this equation.

Using the quadratic formula:

x = (5 ± √((-5)² - 4 * 1 * (-24))) / (2 * 1) x = (5 ± √(25 + 96)) / 2 x = (5 ± √121) / 2 x = (5 ± 11) / 2

There are two solutions:

x₁ = (5 + 11) / 2 = 16/2 = 8 x₂ = (5 - 11) / 2 = -6/2 = -3

So, the solutions are x = 8 and x = -3.

4. 2x^2 + 13x - 6 = 0

a = 2, b = 13, and c = -6 in this equation.

Using the quadratic formula:

x = (-13 ± √(13² - 4 * 2 * (-6))) / (2 * 2) x = (-13 ± √(169 + 48)) / 4 x = (-13 ± √217) / 4

There are two solutions:

x₁ = (-13 + √217) / 4 x₂ = (-13 - √217) / 4

These are the exact solutions. If you want decimal approximations, you can calculate them.

5. 7x^2 - 6x + 2 = 0

This equation doesn't have real solutions because the discriminant (the value inside the square root in the quadratic formula) is negative. So, there are no real solutions.

6. 4x^2 + 12x + 9 = 0

a = 4, b = 12, and c = 9 in this equation.

Using the quadratic formula:

x = (-12 ± √(12² - 4 * 4 * 9)) / (2 * 4) x = (-12 ± √(144 - 144)) / 8 x = (-12 ± √0) / 8

The discriminant is zero, which means there is only one real solution:

x₁ = (-12 + 0) / 8 = -12/8 = -3/2

So, the solution is x = -3/2.