Let's solve the equation 2k^2 = 9 + 3k First, subtract each side by (9+3k) to get 0 on the right side of the equation 2k^2 = 9 + 3k 2k^2 - (9+3k) = 9+3k - (9+3k) 2k^2 - 9 - 3k = 9 + 3k - 9 - 3k 2k^2 - 3k - 9 = 0
As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9. Δ = b^2 - 4ac Δ = (-3)^2 - 4 (2)(-9) Δ<u /> = 9 + 72 Δ<u /> = 81
Δ<u />>0 so the equation got 2 real solutions: k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3 AND k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2
So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2
A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.
3 is an integer number, so it's rational. -3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.
So 2k^2 = 9 + 3k have two rational solutions (Option B).
