中四一條quadratic equation的題目

Math 科方法

Let a and (a+5) be the roots of the given equation.

Put x =a into the equation
2a^2 + 6a + m =0
m = -2a^2 - 6a

Put x =a+5 into the equation
2(a+5)^2+6(a+5)+m=0
2(a+5)^2+6(a+5)+( -2a^2 - 6a )=0 (sub m = -2a^2 - 6a )
2(a^2+10a+25)+6a+30-2a^2 - 6a =0
2a^2+20a+50+6a+30-2a^2-6a=0
20a+80=0
a=-4

sub a = -4 into m = -2a^2 - 6a
m = -2(-4)^2-6(-4)=-2(16)+24=-8


AMath 科方法

Let a and a+5 be the roots of the given equation

sum of roots = a + a+5 = -6/2
2a+5=-3
2a=-8
a=-4

Product of roots = a(a+5)=m/2
-4(-4+5)=m/2
-4(+1)*2 =m
m=-8

as two roots are differ by 5,
then let a root be "a", another root be "a+5"

For eqt. 2X^2+6x+m=0
2a^2+6a+m=0 ---(1)
2(a+5)^2+6(a+5)+m=0 ---(2)
i.e.
2a^2+6a+m=2(a+5)^2+6(a+5)+m = 2a^2+20a+50+6a+50+m = 2a^2+26a+(80+m)
6a=26a+80
-20a=80
a= -4
therefore,
the roots of this eqt. = -4,1

Sub the root x=1 into eqt. 2X^2+6x+m=0
2x1^2 + 6x1 + m =0
m = -8