Question

aplique factorización para resolver ecuaciones cuadratica 7(×+5)(×-3)-5(×+3)(×-3)=0​

Answer 1

Respuesta:

7(x+5)(x-3)-5(x+3)(x-3) = 0​

7((x+5)(x-3))-5((x+3)(x-3)) = 0

7(x²+(5-3)x+(5(-3))-5(x²-3(3)) = 0

7(x²+2x+(-15))-5(x²-9) = 0

7(x²+2x-15)-5x²+45 = 0

7x²+14x-105-5x²+45 = 0

(7-5)x²+14x+45-105 = 0

2x²+14x+(-60) = 0

2x²+14x-60 = 0

2x²+20x-6x-60 = 0

2x(x+10)-6(x+10) = 0

(x+10)(2x-6) = 0

(x+10)(2(x-3)/2) = 0

(x+10)(x-3) = 0

X1 = -10 ; X2 = 3

Explicación paso a paso: