Both expression have the same denominator: 9x²-1. Thus it must not be 0.
9x²-1=(3x-1)(3x+1)=0, resulting x=+-1/3.
Restrictions: x in R\{-1/3, 1/3}
Adding those expressions:
E=(-x-2)/(9x²-1 ) + (-5x+4)/(9x²-1)=
(-x-2-5x+4)/(9x²-1)=(-6x+2)/(9x²-1)=
(-2)(3x-1)/(9x²-1)=-2/(3x+1)
E=-2/(3x+1)
10g^3 - 8g^2 + 5g - 14 + 10g^2 + 12g
its just a matter of combining like terms
10g^3 + 2g^2 + 17g - 14
Use the formula or complete the square.
The zeroes of the quadratic can be real and rational; real and irrational; complex conjugates.
If the quadratic is ax²+bx+c, x=(-b+√b²-4ac)/2a.
If b² > 4ac the solutions are real. If b²-4ac is a perfect square, the solutions are real and rational; otherwise they’re real but irrational.
If b² < 4ac the solutions are complex.
Answer:
The answer is C
Step-by-step explanation:
