Question

1) x2 - 5x + 4 = 0; 3) x² + 4x + 3 = 0; 5) x² + x - 6 = 0;​

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

It seems like you've listed quadratic equations in the form $ax^2 + bx + c = 0$. I can help you solve each of these quadratic equations for $x$.

1. $x^2 - 5x + 4 = 0$: To solve this quadratic equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -5$, and $c = 4$. Plugging these values into the quadratic formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 - 16}}{2}$ $x = \frac{5 \pm \sqrt{9}}{2}$ $x_1 = \frac{5 + 3}{2} = 4$ $x_2 = \frac{5 - 3}{2} = 1$ So the solutions are $x = 4$ and $x = 1$.

3. $x^2 + 4x + 3 = 0$: In this case, $a = 1$, $b = 4$, and $c = 3$. Plugging these values into the quadratic formula: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)}$ $x = \frac{-4 \pm \sqrt{16 - 12}}{2}$ $x = \frac{-4 \pm \sqrt{4}}{2}$ $x_1 = \frac{-4 + 2}{2} = -1$ $x_2 = \frac{-4 - 2}{2} = -3$ So the solutions are $x = -1$ and $x = -3$.

4. $x^2 + x - 6 = 0$: In this case, $a = 1$, $b = 1$, and $c = -6$. Plugging these values into the quadratic formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 24}}{2}$ $x = \frac{-1 \pm \sqrt{25}}{2}$ $x_1 = \frac{-1 + 5}{2} = 2$