Question

1) 9x^2+6x-8=0 2) 25x^2-10•x-3=0 3) 5x^2-7x-6=0 4) x^2-3x-10=0

Answer 1

Ответ:

D=b^2-4ac=6^2-4×9×(-8)=36+288=324

х1=(-6+√324)/9×2=12:18=2/3

х2=(-6-√324)/9×2=-24/18=4/3

Ответ'2/3,4/3

Объяснение:

/-дробь .формула дискриминанта:D=b^2-4ac

a,b,c это коэфицэнты, то есть в данном примере а=9, b=6, c=-8

формула для нахождения корней(x1,x2): для х1=-b+✓D/a×2,а для х2=-b-√D/a×2.

надеюсь ты понял(поняла) остальные примеры решай по формулам и всё получится;)

Answer 2

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

For an equation of the form ax^2 + bx + c = 0, the quadratic formula states that the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's solve each equation one by one:

1. 9x^2 + 6x - 8 = 0

Here, a = 9, b = 6, and c = -8. Plugging these values into the quadratic formula:

x = (-6 ± √(6^2 - 4 * 9 * -8)) / (2 * 9) x = (-6 ± √(36 + 288)) / 18 x = (-6 ± √324) / 18 x = (-6 ± 18) / 18

This gives us two solutions: x₁ = (-6 + 18) / 18 = 12 / 18 = 2 / 3 x₂ = (-6 - 18) / 18 = -24 / 18 = -4 / 3

Therefore, the solutions to the equation 9x^2 + 6x - 8 = 0 are x = 2/3 and x = -4/3.

2. 25x^2 - 10x - 3 = 0

Here, a = 25, b = -10, and c = -3. Plugging these values into the quadratic formula:

x = (-(-10) ± √((-10)^2 - 4 * 25 * -3)) / (2 * 25) x = (10 ± √(100 + 300)) / 50 x = (10 ± √400) / 50 x = (10 ± 20) / 50

This gives us two solutions: x₁ = (10 + 20) / 50 = 30 / 50 = 3 / 5 x₂ = (10 - 20) / 50 = -10 / 50 = -1 / 5

Therefore, the solutions to the equation 25x^2 - 10x - 3 = 0 are x = 3/5 and x = -1/5.

3. 5x^2 - 7x - 6 = 0

Here, a = 5, b = -7, and c = -6. Plugging these values into the quadratic formula:

x = (-(-7) ± √((-7)^2 - 4 * 5 * -6)) / (2 * 5) x = (7 ± √(49 + 120)) / 10 x = (7 ± √169) / 10 x = (7 ± 13) / 10

This gives us two solutions: x₁ = (7 + 13) / 10 = 20 / 10 = 2 x₂ = (7 - 13) / 10 = -6 / 10 = -3 / 5

Therefore, the solutions to the equation 5x^2 - 7x - 6 = 0 are x = 2 and x = -3/5.

4. x^2 - 3x - 10 = 0

Here, a = 1, b = -3, and c = -10. Plugging these values into the quadratic formula:

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

This gives us two solutions: x₁ = (3 + 7) / 2 = 10 / 2 = 5 x₂ = (3 - 7) / 2 = -4 / 2 = -2

Therefore, the solutions to the equation x^2 - 3x - 10 = 0 are x = 5 and x = -2.