Question

Ayuda a resolver las ecuaciones cuadraticas
a) 21x²+100= -5

b) 2x²-6x=6x²-8x

c) (x-3)-(2x+5)² = -16

d) (4x-1) (2x+3) = (x+3) (x-1)

Answer 1

a) 21x²+100= -5;  21X² + 105 ═ 0; Donde a=21;  b =0 ; c = 105

 $X=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$X=\frac{0\pm \sqrt{-4(21)(105)}}{2(21)}$

$X=\frac{0\pm \sqrt{-8820}}{42}$

X1 = +[93.9148 i]/42 = 2.236 i

X2 = -[93.9148 i]/42 = -2.236 i

b) 2x²-6x=6x²-8x;   0 = 6X² - 2X² - 8X + 6X; 0 = 4X² - 2X

Donde a = 4;  b = -2; c= 0

$X=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

 $X=\frac{-(-2)\pm \sqrt{(-2)^2-4(4)(0)}}{2(4)}$

$X=\frac{2\pm \sqrt{4}}{8}$

X1 = [2 + 2]/8 = 1/2

X2 = [2 - 2]/8 = 0

c) (x-3)-(2x+5)² = -16;  (X - 3) - (4X² + 20X + 25) = 16

X - 3 - 4X² - 20X - 25 = -16

-4X² - 19X - 28 = -16

0 = 4X² + 19X + 12 

$X=\frac{-19\pm \sqrt{(19)^2-4(4)(12)}}{2(4)}$

$X=\frac{-19\pm \sqrt{361-192}}{8}$

$X=\frac{-19\pm \sqrt{169}}{8}$

$X=\frac{-19\pm \sqr13}{8}$

X1= [-19+13 ]/8 = -3/4

X2= [-19 - 13]/8 = -4

d) (4x-1) (2x+3) = (x+3) (x-1)
 
(4x-1) (2x+3) = 8X² + 12X - 2X - 3; = 8X² + 10X - 3

 (x+3) (x-1) = X² - X + 3X - 3; X² + 2X - 3

8X² + 10X - 3 = X² + 2X - 3

7X² + 8X = 0;  Donde a = 7;  b = 8,  C = 0

$X=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$X=\frac{-8\pm \sqrt{8^2}}{2(7)}$

X1 = [-8 + 8]/14 = 0

X2 = [-8 - 8]/14 = -16/14 = -8/7