Question

1) x²-5x+4=0 2) 4x²-7x+19=0 3) 5x⁴-7x²-4=0

Answer 1

1) можно не делать подсчеты ,по Виету это числа 1 и 4. Да, запомните, если сумма коэффициентов равна 1 ,то один из корней единица.

2) Дискриминант =49-16*19=49-304=-255 - отрицательный. нет корней.

3) х²=(7±√(49+80))/10=(7±√(129))/10; х²=(7-√(129))/10;∅, т.к. меньше нуля.

х=±√((7+√(129))/10)

Answer 2

It looks like you've provided three quadratic equations. To solve each equation, we can use the quadratic formula, which states that for an equation of the form ax² + bx + c = 0, the solutions for x are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's solve each of your equations using this formula:

1. $x² - 5x + 4 = 0$: Here, a = 1, b = -5, and c = 4. Using the quadratic formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 - 16}}{2}$ $x = \frac{5 \pm \sqrt{9}}{2}$ $x = \frac{5 \pm 3}{2}$

So, the solutions are: $x_1 = \frac{5 + 3}{2} = 4$ $x_2 = \frac{5 - 3}{2} = 1$

2. $4x² - 7x + 19 = 0$: Here, a = 4, b = -7, and c = 19. Using the quadratic formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(4)(19)}}{2(4)}$ $x = \frac{7 \pm \sqrt{49 - 304}}{8}$ $x = \frac{7 \pm \sqrt{-255}}{8}$

Since the discriminant (-255) is negative, the solutions are complex numbers: $x_1 = \frac{7 + i\sqrt{255}}{8}$ $x_2 = \frac{7 - i\sqrt{255}}{8}$

3. $5x⁴ - 7x² - 4 = 0$: This equation is a quartic equation in terms of x². Let's use substitution: Let $y = x²$. Then the equation becomes: $5y² - 7y - 4 = 0$

This is a quadratic equation in terms of y. Let's solve it using the quadratic formula: Here, a = 5, b = -7, and c = -4. $y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(5)(-4)}}{2(5)}$ $y = \frac{7 \pm \sqrt{49 + 80}}{10}$ $y = \frac{7 \pm \sqrt{129}}{10}$

So, the solutions for y are: $y_1 = \frac{7 + \sqrt{129}}{10}$ $y_2 = \frac{7 - \sqrt{129}}{10}$