Question

Розвязок теоремою вієта

Answer 1

По теореме Виета:

${ax}^{2} + bx + c = 0 \\ x_{1} + x _{2} = - \frac{b}{a} \\ x_{1} \times x _{2} = \frac{c}{a}$

3)

1)

$2 {x}^{2} - 3x - 5 = 0 \\ x_{1} + x _{2} = - ( \frac{ - 3}{2}) = 1.5 \\ x_{1} \times x _{2} = \frac{ - 5}{2} = - 2.5\\ x_{1} = - 1\\ x _{2} = 2.5$

2)

${x}^{2} + 2x - 15 = 0 \\ x_{1} + x _{2} = - 2 \\ x_{1} \times x _{2} = - 15 \\ x_{1} = - 5 \\ x _{2} = 3$

3)

$8 {x}^{2} + x - 1 = 0 \\ x_{1} + x _{2} = - \frac{1}{8} \\ x_{1} \times x _{2} = - \frac{1}{8} \\ d = {1}^{2} - 4 \times 8 \times ( - 1) = 1 + 32 = 33 \\ x_{1} = \frac{ - 1 + \sqrt{33} }{16} \\ x _{2} = \frac{ - 1 - \sqrt{33} }{ 16}$

5)

${x}^{2} + bx - 24 = 0 \\ x_{1} = 4 \\ \\ {4}^{2} + 4b - 24 = 0 \\ 16 - 2 4+ 4b = 0 \\ - 8 + 4b = 0 \\ 4b = 8 \\ b = 8 \div 4 \\ b = 2 \\ \\ {x}^{2} + 2x - 24 = 0 \\ x_{1} + x _{2} = - 2\\ x_{1} \times x _{2} = - 24 \\ x_{1} = 4\\ x _{2} = - 6$