Question

1) 4x2-20=0 2) 3x2+5x=0 3) X2-5x-24=0 4) 2x2-13x+6=0 5) 7x2-6x+2=0 6) 4x2+12x+9=0

Answer 1

Формула квадратных уравнений:
Сначала находим коэффициенты:a,b,c
Находим дискриминант(D): D=b^2-4ac
x1=-b+корень от D/2a
x2=-b-корень от D/2a

В пятом там нет корней,следовательно не возможно решать далее

Answer 2

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

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

Here, the equations are in the form $ax^2 + bx + c = 0$.

Let's solve each equation:

1. $4x^2 - 20 = 0$

   $a = 4, b = 0, c = -20$

   $x = \frac{-0 \pm \sqrt{0^2 - 4(4)(-20)}}{2(4)}$

   $x = \frac{\pm \sqrt{320}}{8} = \frac{\pm \sqrt{80}}{2} = \pm \sqrt{80}$

   So, $x = \pm 4 \sqrt{5}$.

2. $3x^2 + 5x = 0$

   $a = 3, b = 5, c = 0$

   $x = \frac{-5 \pm \sqrt{5^2 - 4(3)(0)}}{2(3)}$

   $x = \frac{-5 \pm \sqrt{25}}{6} = \frac{-5 \pm 5}{6}$

   So, $x = \frac{-10}{6} = -\frac{5}{3}$ or $x = 0$.

3. $x^2 - 5x - 24 = 0$

   $a = 1, b = -5, c = -24$

   $x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-24)}}{2(1)}$

   $x = \frac{5 \pm \sqrt{121}}{2} = \frac{5 \pm 11}{2}$

   So, $x = 8$ or $x = -3$.

4. $2x^2 - 13x + 6 = 0$

   $a = 2, b = -13, c = 6$

   $x = \frac{13 \pm \sqrt{(-13)^2 - 4(2)(6)}}{2(2)}$

   $x = \frac{13 \pm \sqrt{169 - 48}}{4} = \frac{13 \pm \sqrt{121}}{4}$