Równanie kwadratowe zupełne. Proszę o rozwiązania ;)
a) x²-10x+25=0 b) 4x²=12x-9
c) 9x²+1=6x d) 16x²+25=40x
e) 9x²=12x-4 f) 25x²+10x+1=0


Odpowiedź :

Odpowiedź:

[tex]a) {x}^{2} - 10x + 25 = 0 \\ (x - 5 {)}^{2} = 0 \\ x - 5 = 0 \\ x = 5[/tex]

[tex]b)4 {x}^{2} = 12x - 9 \\ 4 {x}^{2} - 12x + 9 = 0 \\ (2x - 3 {)}^{2} = 0 \\ 2x - 3 = 0 \\ 2x = 3 \\ x = \frac{3}{2} [/tex]

[tex]c)9 {x}^{2} + 1 = 6x \\ 9 {x}^{2} - 6x + 1 = 0 \\ (3x - 1 {)}^{2} = 0 \\ 3x - 1 = 0 \\ 3x = 1 \\ x = \frac{1}{3} [/tex]

[tex]d)16 {x}^{2} + 25 = 40x \\ 16 {x}^{2} - 40x + 25 = 0 \\ (4x - 5 {)}^{2} = 0 \\ 4x - 5 = 0 \\ 4x = 5 \\ x = \frac{5}{4} [/tex]

[tex]e)9 {x}^{2} = 12x - 4 \\ 9 {x}^{2} - 12x + 4 = 0 \\ (3x - 2 {)}^{2} = 0 \\ 3x - 2 = 0 \\ 3x = 2 \\ x = \frac{2}{3} [/tex]

[tex]f)25 {x}^{2} + 10x + 1 = 0 \\ (5x + 1 {)}^{2} = 0 \\ 5x + 1 = 0 \\ 5x = - 1 \\ x = - \frac{1}{5} [/tex]