Question

1. x² +11x=0 2. 4x²-2x=0 3. -3x²+7x=0 4. x²-16=0 5. 16x²-9=0 6. x² +25=0​

Answer 1

1) x(x+11)=0 x1=0, x2=-11

2)x(4x-2)=0 x1=0 4x-2=0 x2=½

3)-x(3x-7)=0 x1=0 3x-7=0 x2=7/3

4)x²=16 x1=4 x2=-4

5) 16x²=9 x1=3/4 x2=-3/4

6) x²≠-25 корень не может быть отрицательным

Answer 2

To solve these quadratic equations, we'll use the quadratic formula, which is given by:

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

In each equation, we have $ax^2 + bx + c = 0$, so we'll use this formula to find the solutions for each equation.

1. $x^2 + 11x = 0$ $a = 1$, $b = 11$, $c = 0$ Plug these values into the formula: $x = \frac{-11 \pm \sqrt{11^2 - 4(1)(0)}}{2(1)}$ $x = \frac{-11 \pm \sqrt{121}}{2}$ $x = \frac{-11 \pm 11}{2}$ So, the solutions are $x = -11/2$ and $x = 0$.

2. $4x^2 - 2x = 0$ $a = 4$, $b = -2$, $c = 0$ Plug these values into the formula: $x = \frac{2 \pm \sqrt{(-2)^2 - 4(4)(0)}}{2(4)}$ $x = \frac{2 \pm \sqrt{4}}{8}$ $x = \frac{2 \pm 2}{8}$ So, the solutions are $x = 1/2$ and $x = 0$.

3. $-3x^2 + 7x = 0$ $a = -3$, $b = 7$, $c = 0$ Plug these values into the formula: $x = \frac{-7 \pm \sqrt{7^2 - 4(-3)(0)}}{2(-3)}$ $x = \frac{-7 \pm \sqrt{49}}{-6}$ $x = \frac{-7 \pm 7}{-6}$ So, the solutions are $x = 1$ and $x = 0$.

4. $x^2 - 16 = 0$ $a = 1$, $b = 0$, $c = -16$ Plug these values into the formula: $x = \frac{0 \pm \sqrt{0^2 - 4(1)(-16)}}{2(1)}$ $x = \frac{\pm \sqrt{64}}{2}$ $x = \frac{\pm 8}{2}$ So, the solutions are $x = 4$ and $x = -4$.

5. $16x^2 - 9 = 0$ $a = 16$, 0 0