Write a quadratic equation that has the given solutions.
1. 3 and 2
2.1 and 3
2007-07-02 07:54:16 · 4 answers · asked by Berry A 2 in Science & Mathematics ➔ Mathematics
1. 3 and 2
(x-3)(x-2)
so x^2 -3x-2x+6
=x^2 -5x+6
2.1 and 3
(x-1)(x-3)
x^2-x-3x+3
=x^2 -4x+3
2007-07-02 07:57:06 · answer #1 · answered by sweet n simple 5 · 2⤊ 0⤋
If x = 3 is a root, then x-3 is a factor.
If x = 2 is a root, then x-2 is a factor.
Thus (x-3)(x-2)=0 has roots 3 and 2, so
x^2 - 5x + 6 = 0 is your equation.
Same for 1 and 3, so
(x-1)(x-3)=0 is the same as
x^2 - 4x + 3 = 0.
2007-07-02 14:58:43 · answer #2 · answered by MathProf 4 · 0⤊ 0⤋
1.
x=3 or x=2
x-3 = 0 or x-2 = 0
(x-3)(x-2) = 0
x^2 - 2x - 3x +6 = 0
x^2 - 5x + 6 =0
2.
similarly (x-3)(x-1) = 0
x^2 - x - 3x + 3 = 0
x^2 - 4x + 3 =0
2007-07-02 14:59:31 · answer #3 · answered by Vipin A 3 · 0⤊ 0⤋
1. x^2- 5x+6
2. x^2- 4x+3
2007-07-02 15:00:15 · answer #4 · answered by Archmage 2 · 0⤊ 0⤋