a) x(4^2 + 12x + 9) = 0 b) x(2x^2 - 5x + 3) = 0

Odpowiedź :

a)

[tex]x(4 {x}^{2} + 12x + 9) = 0[/tex]

[tex]x = 0[/tex]

[tex]4 {x}^{2} + 12x + 9 = 0[/tex]

∆=

[tex] {b}^{2} - 4ac = {12}^{2} - 4 \times 4 \times 9 = 144 - 144 = 0[/tex]

[tex] x_{0} = \frac{ - b}{2a} = \frac{ - 12}{2 \times 4} = \frac{ - 12}{8} = - 1 \frac{4}{8} = - 1 \frac{1}{2} [/tex]

b)

[tex]x(2 {x}^{2} - 5x + 3) = 0[/tex]

[tex]x = 0[/tex]

[tex]2 {x}^{2} - 5x + 3 = 0[/tex]

∆=

[tex] { (- 5)}^{2} - 4 \times 2 \times 3 = 25 - 24 = 1[/tex]

√∆=1

[tex] x_{1} = \frac{ - b - \sqrt{delty} }{2a} = \frac{ - ( - 5) - 1}{2 \times 2} = \frac{5 - 1}{4} = \frac{4}{4} = 1[/tex]

[tex] x_{2} = \frac{ - b + \sqrt{delty} }{2a} = \frac{ - ( - 5) + 1}{2 \times 2} = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2} = 1 \frac{1}{2} [/tex]