If the function f x bx a b is not 0 and there is a zero point 3, find the zero point of the function

Solution: Derived from the question.

f(-3)=-3b+a=0

a=3bg(x)=3bx^2+2bx

bx(3x+2)

Find the zero point of g(x).

Let g(x)=0

g(x)=bx(3x+2)=0

Solution: x=0, or x=-2 3

The zero points of the function g(x) are 0 and -2 3

Because f(x) has a zero point that is -3, it is brought in to get it.

0=-3b+a, i.e. a=3b

The zero point of g(x)=x(ax+2b) is 0 or -2b a, which is obtained because a==3b.

The zero point of g(x) is 0 and -2 3

The function f(x)=ax+b has a zero point of 2

i.e. 2a+b=0

So b=-2a

g(x)=bx 2-ax=-2ax 2-ax=-a(2x 2-x), so the two zeros are 0 and 1 2

The former has a zero point of 2, then 2a+b=0 b=-2ag(x)=x(-2ax-a)=-ax(2x+1) because a≠0

So g(x) two zeros are (0,0)(-1 2,0).

According to the title, there are: ax+b=0(a≠0)x=-b a.Another midnight 2.

b a=2, i.e. b = -2aAnd g(x)=bx-ax=0, that is, bx -ax=x(bx-a)=x(-2ax-a)=-ax(2x+1)=0Here a ≠ 0, so the zero point has two 0s, -1 2.

Because there is a zero point 3, so 3a-b=0, so b=3a, substituting b=3a into g(x)=bx2+3ax, we get 3ax 2+3ax=0, and the zero point is -1,0

Because f(x) has a zero point that is 2

So the substitution gets:

2a - b = 0

So b=-2a

g(x) = 0

bx² +ax = 0

2ax² +ax = 0

2x² +x = 0

x(2x + 1) = 0

x = 0 or x = -1 2

So the zero point of g(x) is 0 and -1 2

The function f(x)=ax-b(a≠0) has a zero point of 2, over (2,0), 0=2a-b, 2a=b

g(x)=bx 2+ax =2ax 2+ax (a≠0), let g(x)=0 ,2x 2+x=0 ,x(2x+1)=0 ,x=0 ,x=-1 2

Substituting the zero point is 2.

2a+b=0

b= -2a

g(x)= 2ax^2 -ax

Substituting g(x)=0 is obtained.

2ax^2 -ax=0

Because a≠0 is eliminated, a is obtained.

2x^2 +x=0

The zero point is minus half or zero.

f(x)=ax+b (a is not equal to 0) has a zero point which is 22a+b=0 b=-2a

g（x）=bx²-ax＝0

x﹙bx-a)=0

x=0 x=a/b=-1/2

The zero points of the function g(x)=bx 2-ax are 0 and -1 2