Question

A) x²+x-6=0 b) x²– 12x + 27 = 0c) y²– 8y - 20 = 0d) x² + 17x + 16 = 0решите плиз 13 баллов .​

Answer 1

Ответ:

Пошаговое объяснение:

а)

х²+х-6=0.

х₁₂= (-1±√(1+4*6))/2=( -1±5)/2.

х₁ =(-1+5)/2=4/2=2.

х₂=(-1-5)/2=-3.

б)

х²-12х+27=0.

х₁₂=6±√(36-27)=6±3.

х₁=9.

х₂=3.

Answer 2

To solve each quadratic equation, we can use the quadratic formula, which states that for a quadratic equation of the form ax² + bx + c = 0, the solutions for x can be found using:

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

Let's solve each equation step by step:

A) $x² + x - 6 = 0$

In this equation, a = 1, b = 1, and c = -6. Plugging these values into the quadratic formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}$

$x = \frac{-1 \pm \sqrt{1 + 24}}{2}$

$x = \frac{-1 \pm \sqrt{25}}{2}$

$x = \frac{-1 \pm 5}{2}$

So, the solutions for A) are:

$x_1 = \frac{-1 + 5}{2} = 2$

$x_2 = \frac{-1 - 5}{2} = -3$

B) $x² - 12x + 27 = 0$

In this equation, a = 1, b = -12, and c = 27. Plugging these values into the quadratic formula:

$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1}$

$x = \frac{12 \pm \sqrt{144 - 108}}{2}$

$x = \frac{12 \pm \sqrt{36}}{2}$

$x = \frac{12 \pm 6}{2}$

So, the solutions for B) are:

$x_1 = \frac{12 + 6}{2} = 9$

$x_2 = \frac{12 - 6}{2} = 3$

C) $y² - 8y - 20 = 0$

In this equation, a = 1, b = -8, and c = -20. Plugging these values into the quadratic formula:

$y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1}$

$y = \frac{8 \pm \sqrt{64 + 80}}{2}$

$y = \frac{8 \pm \sqrt{144}}{2}$

$y = \frac{8 \pm 12}{2}$

So, the solutions for C) are:

$y_1 = \frac{8 + 12}{2} = 10$

$y_2 = \frac{8 - 12}{2} = -2$

D) $x² + 17x + 16 = 0$

In this equation, a = 1, b = 17, and c = 16. Plugging these values into the quadratic formula:

0 0