8. Решите уравнение, используя формулу корней квадратного уравнения, у которого второй коэффициент является четным числом:

Решение:

Для уравнений вида $$
ax^2+2kx+c=0
корни можно найти по формуле:
$$
x_{1,2} = \(\frac{-k \pm \sqrt{k^2-ac}}{a}\)

1) 5x² - 14x - 3 = 0

Здесь $$
a=5, k=-7, c=-3

$$
x_{1,2} = \(\frac{-(-7) \pm \sqrt{(-7)^2-5(-3)}}{5}\) = \(\frac{7 \pm \sqrt{49+15}}{5}\) = \(\frac{7 \pm \sqrt{64}}{5}\) = \(\frac{7 \pm 8}{5}\)

$$
x_1 = \(\frac{7+8}{5}\) = \(\frac{15}{5}\) = 3

$$
x_2 = \(\frac{7-8}{5}\) = \(\frac{-1}{5}\)

3) 4x² - 8x + 3 = 0

Здесь $$
a=4, k=-4, c=3

$$
x_{1,2} = \(\frac{-(-4) \pm \sqrt{(-4)^2-4(3)}}{4}\) = \(\frac{4 \pm \sqrt{16-12}}{4}\) = \(\frac{4 \pm \sqrt{4}}{4}\) = \(\frac{4 \pm 2}{4}\)

$$
x_1 = \(\frac{4+2}{4}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)

$$
x_2 = \(\frac{4-2}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

5) 9x² - 24x - 20 = 0

Здесь $$
a=9, k=-12, c=-20

$$
x_{1,2} = \(\frac{-(-12) \pm \sqrt{(-12)^2-9(-20)}}{9}\) = \(\frac{12 \pm \sqrt{144+180}}{9}\) = \(\frac{12 \pm \sqrt{324}}{9}\) = \(\frac{12 \pm 18}{9}\)

$$
x_1 = \(\frac{12+18}{9}\) = \(\frac{30}{9}\) = \(\frac{10}{3}\)

$$
x_2 = \(\frac{12-18}{9}\) = \(\frac{-6}{9}\) = \(\frac{-2}{3}\)

2) 15x² - 2x - 1 = 0

Здесь $$
a=15, k=-1, c=-1

$$
x_{1,2} = \(\frac{-(-1) \pm \sqrt{(-1)^2-15(-1)}}{15}\) = \(\frac{1 \pm \sqrt{1+15}}{15}\) = \(\frac{1 \pm \sqrt{16}}{15}\) = \(\frac{1 \pm 4}{15}\)

$$
x_1 = \(\frac{1+4}{15}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\)

$$
x_2 = \(\frac{1-4}{15}\) = \(\frac{-3}{15}\) = \(\frac{-1}{5}\)

4) 5x² - 6x + 1 = 0

Здесь $$
a=5, k=-3, c=1

$$
x_{1,2} = \(\frac{-(-3) \pm \sqrt{(-3)^2-5(1)}}{5}\) = \(\frac{3 \pm \sqrt{9-5}}{5}\) = \(\frac{3 \pm \sqrt{4}}{5}\) = \(\frac{3 \pm 2}{5}\)

$$
x_1 = \(\frac{3+2}{5}\) = \(\frac{5}{5}\) = 1

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

6) 4x² - 4x - 15 = 0

Здесь $$
a=4, k=-2, c=-15

$$
x_{1,2} = \(\frac{-(-2) \pm \sqrt{(-2)^2-4(-15)}}{4}\) = \(\frac{2 \pm \sqrt{4+60}}{4}\) = \(\frac{2 \pm \sqrt{64}}{4}\) = \(\frac{2 \pm 8}{4}\)

$$
x_1 = \(\frac{2+8}{4}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\)

$$
x_2 = \(\frac{2-8}{4}\) = \(\frac{-6}{4}\) = \(\frac{-3}{2}\)