bardzo proszę o pomoc w rozwiązaniu zadania. daje naj​



Bardzo Proszę O Pomoc W Rozwiązaniu Zadania Daje Naj class=

Odpowiedź :

[tex]ax^{2}+bx + c = 0\\\\\Delta = b^{2}-4ac\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a}[/tex]

2.81.

[tex]a)\\x^{2}+4x-5 = 0\\\\\Delta = 4^{2}-4\cdot1\cdot(-5) = 16 + 20 = 36\\\\\sqrt{\Delta} = \sqrt{36} = 6\\\\x_1 = \frac{-4-6}{2} = \frac{-10}{2} = -5\\\\x_2 = \frac{-4+6}{2} = \frac{4}{2} = 2[/tex]

[tex]b)\\-x^{2}+4x+21 = 0\\\\\Delta = 4^{2}-4\cdot(-1)\cdot21 = 16 +84 = 100\\\\\sqrt{\Delta} = \sqrt{100} = 10\\\\x_1 = \frac{-4+10}{-2} = \frac{6}{-2} = -3\\\\x_2 = \frac{-4-10}{-2} = \frac{-14}{-2} = 7[/tex]

2.82.

a)

2x² + 9x - 35 = 0

Δ = 9² - 4·2·(-35) = 81 + 280 = 361

√Δ = √361 = 19

x₁ = (-9 - 19)/4 = -28/4 = -7

x₂ = (-9 + 19)/4 = 10/4 = 2,5

b)

-x² + 5x - 15 = 0

Δ = 5² - 4·(-1)·(-15) = 25 - 60 = -35 < 0

Δ < 0, brak pierwiastków równania