Question

2) 7x2 + 18x + 5 = 0; 4) 9x2 - 4x - 2 = 0.​

Answer 1

Ответ:

Четвертое лень писать,надеюсь что нужно было через дискриминант

Answer 2

It looks like you've provided two quadratic equations. To solve them, we'll use the quadratic formula:

For an equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

Let's apply this formula to each of the equations:

1. $7x^2 + 18x + 5 = 0$:

   Here, $a = 7$, $b = 18$, and $c = 5$.

   Plugging these values into the formula:

   $x = \frac{{-18 \pm \sqrt{{18^2 - 4 \cdot 7 \cdot 5}}}}{{2 \cdot 7}}$

   $x = \frac{{-18 \pm \sqrt{{324 - 140}}}}{{14}}$

   $x = \frac{{-18 \pm \sqrt{{184}}}}{{14}}$

   $x = \frac{{-18 \pm 2\sqrt{{46}}}}{{14}}$

   Simplifying further:

   $x = \frac{{-9 \pm \sqrt{{46}}}}{{7}}$

   So, the solutions for the first equation are $x = \frac{{-9 + \sqrt{{46}}}}{{7}}$ and $x = \frac{{-9 - \sqrt{{46}}}}{{7}}$.

2. $9x^2 - 4x - 2 = 0$:

   Here, $a = 9$, $b = -4$, and $c = -2$.

   Plugging these values into the formula:

   $x = \frac{{4 \pm \sqrt{{(-4)^2 - 4 \cdot 9 \cdot (-2)}}}}{{2 \cdot 9}}$

   $x = \frac{{4 \pm \sqrt{{16 + 72}}}}{{18}}$

   $x = \frac{{4 \pm \sqrt{{88}}}}{{18}}$

   $x = \frac{{2 \pm \sqrt{{22}}}}{{9}}$

   So, the solutions for the second equation are $x = \frac{{2 + \sqrt{{22}}}}{{9}}$ and $x = \frac{{2 - \sqrt{{22}}}}{{9}}$.