一元二次方程問題

x[(2k-1)x+3](3k+1)+9=0
so,( 2 k - 1 ) x^2 + 3 ( 3 k + 4 ) =0
有2個相等實數根
→ - 4 ( 2 k - 1) 3 ( 3 k + 4 )=0
6 k^2 +5 k - 4 = 0
( 2 k - 1 ) ( 3 k + 4) = 0
咁 k = 1/2 OR -4/3

x[(2k-1)x+3](3k+1)+9=0
(2kx^2-x^2+3x)(3k+1)+9=0

COZ IT HAS 2 EQUAL ROOTS,
SO, D=0
(9k+3)^2-4(5k^2+2k-1)(9)=0
-99k^2+126k-27=0
k=3/11 OR k=1

SUB. k=3/11 OR k=1 INTO THE EQUATION,
5(3/11 )^2x^2+9x(3/11 )+2(3/11 )x^2-x^2+3x+9=0
-10/121x^2+60/11x+9=0
x=-1.16 or x=67.6(3sig. fig.)
OR
5(1 )^2x^2+9x(1)+2(1)x^2-x^2+3x+9=0
6^2+12+9=0
x= no solution

therefore, x=-1.16 or x=67.6(3sig. fig.)