§ 7. Квадратное уравнение и его корни — 23. Теорема Виета — 580 — стр. 135

Решение квадратного уравнения \(x^2 - 15x - 16 = 0:\)

\(x^2 - 15x - 16 = 0\)

\(x^2 - 16x + x - 16 = 0\)

\(x(x - 16) + (x - 16) = 0\)

\((x + 1)(x - 16) = 0\)

\(\begin{cases}x_1 = -1\\x_2 = 16\end{cases}\)

Проверка:

\(a = 1\).

\(b=-(x_1+x_2)=-(-1+16)=-15\)

\(c=x_1x_2=-1\cdot16=-16\).

Решение \(m^2 - 6m - 11 = 0:\)

\(m^2 - 6m - 11 = 0\)

\(D_1 = (-3)^2 - 1 \cdot (-11) = 9 + 11 = 20\)

\(m_{1,2} = 3 \pm \sqrt{20} = 3 \pm 2 \sqrt{5}\)

Проверка:

\(a = 1\)

\(b = - (x_1 + x_2) = - (3 - 2 \sqrt{5} + 3 + 2 \sqrt{5}) = -6 \)

\(c = x_1 \cdot x_2 = (3 - 2 \sqrt{5}) \cdot (3 + 2 \sqrt{5}) = 9 - 20 = -11\).

Решение \(12 x^{2}-4 x-1=0:\)

\(12 x^{2}-4 x-1=0\)

\(D_1 = (-2)^2 - 12 \cdot (-1) = 4 + 12 = 16 = 4^2\)

\(x = \frac{2 \pm 4}{12}, \quad \begin{cases}x_1 = -\frac{1}{6} \\x_2 = \frac{1}{2}\end{cases}\)

Проверка:

\(a = 12\)

\(\frac{b}{a} = - (x_1 + x_2) = - ( -\frac{1}{6} + \frac{1}{2} ) = -\frac{1}{3} = \frac{-4}{12}\)

\(\frac{c}{a} = x_1 \cdot x_2 = -\frac{1}{6} \cdot \frac{1}{2} = \frac{-1}{12} \).

Решение \(t^2 - 6 = 0 :\)

\(t^2 - 6 = 0 \)

\( t^2 = 6\)

\( t_{1,2} = \pm \sqrt{6}\)

Проверка

\(a = 1:\)

\(b = - (x_1 + x_2) = - (-\sqrt{6} + \sqrt{6}) = 0 \)

\(c = x_1 \cdot x_2 = -\sqrt{6} \cdot \sqrt{6} = -6 \).

Решение \(5x^2 - 18x = 0:\)

\(5x^2 - 18x = 0 \)

\( x(5x - 18) = 0\)

\( \begin{cases}x_1 = 0 \\x_2 = \frac{18}{5} = 3.6\end{cases}\)

Проверка:

\(a = 5:\)

\(\frac{b}{a} = - (x_1 + x_2)= - (0 + 3.6) = -3.6 = \frac{-18}{5}\)

\(\frac{c}{a} = x_1 \cdot x_2 = 0 \cdot 3.6 = 0\).

Решение \(2y^2 - 41 = 0:\)

\(y^2 = \frac{41}{2} = 20.5 \)

\(y_{1,2} = \pm \sqrt{20.5}\)

Проверка:

\(a = 2\)

\(\frac{b}{a} = - (x_1 + x_2) = - (-\sqrt{20.5} + \sqrt{20.5}) = 0 = \frac{0}{2}\)

\(\frac{c}{a} = x_1 \cdot x_2 = -\sqrt{20.5} \cdot \sqrt{20.5} = -20.5 = \frac{-41}{2}\).