Question

1) 4х 2 -20=0; 2) 3х 2 +5х=0; 3) 7х 2 -22х+3=0 4) х 4 -35х 2 -36=0;

Answer 1

Ответ:

А что нужно сделать???

Answer 2

It looks like you've provided four different quadratic equations. To help you, I'll provide solutions for each equation:

1. $4x^2 - 20 = 0$: To solve this quadratic equation, you can first add 20 to both sides and then divide both sides by 4: $4x^2 = 20$ $x^2 = 5$ $x = \pm \sqrt{5}$

So, the solutions are $x = \sqrt{5}$ and $x = -\sqrt{5}$.

2. $3x^2 + 5x = 0$: Here, you can factor out $x$ from both terms: $x(3x + 5) = 0$ This equation is satisfied when either $x = 0$ or $3x + 5 = 0$. Solving the second part: $3x = -5$ $x = -\frac{5}{3}$

So, the solutions are $x = 0$ and $x = -\frac{5}{3}$.

3. $7x^2 - 22x + 3 = 0$: This quadratic equation doesn't factor easily, so you can use the quadratic formula to find the solutions: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For this equation, $a = 7$, $b = -22$, and $c = 3$. Plugging in these values: $x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 7 \cdot 3}}{2 \cdot 7}$ $x = \frac{22 \pm \sqrt{484 - 84}}{14}$ $x = \frac{22 \pm \sqrt{400}}{14}$ $x = \frac{22 \pm 20}{14}$

So, the solutions are $x = \frac{42}{14} = 3$ and $x = \frac{2}{14} = \frac{1}{7}$.

4. $x^4 - 35x^2 - 36 = 0$: This is a quadratic equation in terms of $x^2$. Let $y = x^2$, then the equation becomes: $y^2 - 35y - 36 = 0$

This can be factored: $(y - 36)(y + 1) = 0$

So, $y = 36$ or $y = -1$. Substituting $y = x^2$: $x^2 = 36$ $x = \pm 6$

$x^2 = -1$ This equation has no real solutions since the square of a real number cannot be negative.

Therefore, the solutions are $x = 6$ and $x = -6$.