Question

solve the following quadratic equations

1.300/x-300/x+10=1

2.x-1/x=1

3.x^{2} +[3x+3]^{2} =[3x+4]^{2]

Answer 1

Step-by-step explanation:

1. given equation is not a quadratic equation

2. x-1/x=1

= x^2-1/x=1

= x^2-1=x

= x^2-x-1=0

the answer is in the form of ax^2+bx+c=0

answer is x^2-x-1=0

3. given

x^2+(3x+3)^2=(3x+4)^2

x^2+ 9x^2+9+18x=9x^2+16+24x

x^2-6x-7=0