Question

4х²-3х-1=0; -2х²+х+15=0

Answer 1

$1) \: 4 {x}^{2} - 3x - 1 = 0$
D = в^2-4ac

$d = ( - 3)^{2} - 4 \times 4 \times ( - 1) = 9 - 16 \times ( - 1) = 9 - ( - 16) = 9 + 16 = 25$

D>0



$\sqrt{d } = \sqrt{25} = 5$


Два корня.

$x = \frac{ - b + \sqrt{d} }{2a}$

$x{1} = \frac{3 - 5}{2 \times 4} = - \frac{2}{8} = - \frac{1}{4}$


$x2 = \frac{3 + 5}{2 \times 4} = \frac{8}{8} = 1$


$otvet : x1 = - \frac{1}{4} .x2 = 1$






$2) \: - 2x ^{2} + x + 15 = 0 \\ \\ \: \: - 2 {x}^{2} + x + 15 = 0 | \times ( - 1) \\ \\ \: \: 2 {x}^{2} - x - 15 = 0$


D= в^2-4ас

$d = ( - 1)^{2} - 4 \times 2 \times ( - 15) = 1 - 8 \times ( - 15) = 1 + 120 = 121$
D>0

$\sqrt{d} = \sqrt{121} = 11$

Два корня.

$x = \frac{ - b + \sqrt{d} }{2a}$


$x1 = \frac{1 - 11}{2 \times 2} = - \frac{10}{4} = - \frac{5}{2}$

$x2 = \frac{1 + 11}{2 \times 2} = \frac{12}{4} = 3$

$otvet : x1 = - \frac{5}{2} .x2 = 3$



Вот я помогла