3x^2 = 2x + 5
please leave answers in simplest radical form.
2007-08-02 07:33:44 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3x^2 = 2x + 5
3x^2 - 2x - 5 = 0
(3x - 5)(x + 1) = 0
I rewrote this as a polynomial set equal to zero, and then factored. There is one solution for each factor, where the factor is set equal to zero.
3x - 5 = 0 ==> x = 5/3
x + 1 = 0 ==> x = -1
2007-08-02 07:36:39 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
3x^2 - 2x - 5 = 0, so (3x -5)(x + 1) = 0
So 3x - 5 = 0 and/or x+1 = 0, so
x = 5/3 and x = -1
2007-08-02 07:38:39 · answer #2 · answered by John V 6 · 0⤊ 0⤋
x=5/3
x= -1
2007-08-02 07:38:09 · answer #3 · answered by ]]--((Iam-HeRe))--[[ 2 · 0⤊ 0⤋
3x^2 =2x+5
3x^2-2x-5=0
3x^2 -5x+3x-5=0
x(3x-5)+1(3x-5)=0
(3x-5)(x+1)=0
3x=5, x= -1
x=5/3 , x= -1
2007-08-02 07:41:57 · answer #4 · answered by flori 4 · 0⤊ 0⤋