Question

Решите уравнение: 1) 3x^2 − 15 = 0; 3) x^2 + 8x − 9 = 0; 5) x^2 − 6x − 3 = 0; 2) 4x^2 − 7x = 0; 4) 12x^2 − 5x − 2 = 0; 6) x^2 − 3x + 11 = 0.

Answer 1

1) 3x² - 15 = 0

3x² = 15

x² = 5

x₁ = - √ 5       x₂ = √5

2) 4x² - 7x = 0

x(4x - 7) = 0

или x₁ = 0     или     4x - 7 = 0

                                4x = 7

                                x₂ = 1,75

3) x² + 8x - 9 = 0

По теореме Виета : x₁ + x₂ = - 8        x₁ *x₂ = - 9 , значит

x₁ = - 9            x₂ = 1

4) 12x² - 5x - 2 = 0

D = (- 5)² - 4 * 12 * (- 2) = 25 + 96 = 121 = 11²

$x_{1}=\frac{5+11}{24}=\frac{16}{24}=\frac{2}{3}\\\\x_{2}=\frac{5-11}{24}=-\frac{6}{24}=-0,25$

5) x² - 6x - 3 = 0

D = (- 6)² - 4 * (- 3) = 36 + 12 = 48 = (4√3)²

$x_{1}=\frac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\\\\x_{2}=\frac{6+4\sqrt{3} }{2}=3+2\sqrt{3}$

6) x² - 3x + 11 = 0

D = (- 3)² - 4 * 11 = 9 - 44 < 0

решений нет

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 is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's solve each equation one by one:

1. 3x^2 − 15 = 0

a = 3, b = 0, c = -15

x = (0 ± √(0^2 - 4 * 3 * -15)) / 2 * 3 x = (±√180) / 6 x = (±6√5) / 6 x = ±√5

So the solutions are x = √5 and x = -√5.

2. 4x^2 − 7x = 0

a = 4, b = -7, c = 0

x = (7 ± √((-7)^2 - 4 * 4 * 0)) / 2 * 4 x = (7 ± √(49)) / 8 x = (7 ± 7) / 8

For x = (7 + 7) / 8, x = 14 / 8 = 7 / 4 For x = (7 - 7) / 8, x = 0

So the solutions are x = 7/4 and x = 0.

3. x^2 + 8x − 9 = 0

a = 1, b = 8, c = -9

x = (-8 ± √(8^2 - 4 * 1 * -9)) / 2 * 1 x = (-8 ± √(64 + 36)) / 2 x = (-8 ± √100) / 2 x = (-8 ± 10) / 2

For x = (-8 + 10) / 2, x = 2 / 2 = 1 For x = (-8 - 10) / 2, x = -18 / 2 = -9

So the solutions are x = 1 and x = -9.

4. 12x^2 − 5x − 2 = 0

a = 12, b = -5, c = -2

x = (5 ± √((-5)^2 - 4 * 12 * -2)) / 2 * 12 x = (5 ± √(25 + 96)) / 24 x = (5 ± √121) / 24 x = (5 ± 11) / 24

For x = (5 + 11) / 24, x = 16 / 24 = 2 / 3 For x = (5 - 11) / 24, x = -6 / 24 = -1 / 4

So the solutions are x = 2/3 and x = -1/4.

5. x^2 − 6x − 3 = 0

a = 1, b = -6, c = -3

x = (6 ± √((-6)^2 - 4 * 1 * -3)) / 2 * 1 x = (6 ± √(36 + 12)) / 2 x = (6 ± √48) / 2 x = (6 ± 4√3) / 2 x = 3 ± 2√3

So the solutions are x = 3 + 2√3 and x = 3 - 2√3.

6. x^2 − 3x + 11 = 0

a = 1, b = -3, c = 11

x = (3 ± √((-3)^2 - 4 * 1 * 11)) / 2 * 1 x = (3 ± √(9 - 44)) / 2 x = (3 ± √(-35)) / 2

Since the discriminant (b^2 - 4ac) is negative, the solutions involve complex numbers. We can write them in the form:

x = (3 + √35i) / 2 and x = (3 - √35i) / 2, where i is the imaginary unit (i^2 = -1).

So the solutions are x = (3 + √35i) / 2 and x = (3 - √35i) / 2.