Question about the absolute value equation x2 ax 4

Solution: First of all, remove the absolute value, then there is x 2+ax=4 or x 2+ax=-4, which is two unary quadratic equations x 2+ax-4=0 or x 2+ax+4=0 Since the equation has only three unequal real roots, then there must be an equation with two equal real roots, one square.

Cheng has two unequal real roots.

Obviously, for the equation , 1=a 2+16 0 is constant, indicating that the equation has two unequal real roots. And the equation

There must be two equal real roots.

That is, there is: 2=a 2-16=0 to get a=4 or -4 When a=4, the three roots are: -2-2 2, -2+2 2, -2 When a=-4, the three roots are: 2-2 2, 2+2 2, 2

x2+ax|=4

x 2+ax=4, or x 2+ax=-4

That is, x 2 + ax-4 = 0 or x 2 + ax + 4 = 0 According to the above two equations, one has two different solutions, and the other has only one solution.

Due to the discriminant formula of the first equation.

a 2 + 16 0, it has two different solutions, so it can only be the latter equation x 2 + ax + 4 = 0 has a solution, that is, its discriminant formula δ = a 2-16 = 0, the solution is a = 4 and a = 4 is substituted into the above two equations respectively to get :

x 2 + 4 x - 4 = 0 or x 2 + 4 + 4 = 0 The solution of the first equation is x1=-2-2 2, x2=-2+2 2, and the solution of the second equation is x3=-2

a=-4 is solved in the same way. Complete.

Absolute value of x+4 - absolute value of (x-2) = x+1 when x>2:

x+4-(x-2)=x+1

x+4-x+2=x+1

6 = x + 1x = 5 when - 4

Summary. So 2x+1=4-x or 2x+1+4-x 0 (the sum of two numbers that are opposite to each other is 0).

3.Solve the following absolute value equation: |2x+1|-|4-x||2x+1|-|4-x|What's <>

Hiss. It should be |2x+1|=|4-x|

2x+1|-|4-x|4. Is that so.

Misentered. The absolute value of two numbers is equal, and the two numbers are either equal or opposite to each other.

So 2x+1=4-x or 2x+1+4-x 0 (the sum of two numbers that are opposite to each other is 0).

So you can solve x 1 or x -5

You see, I say that, do you understand<>

<>Oh! Ok Ok.

Got it. Uh-huh.

2x-5|-|4x+7|> beam scatter oak macro = 0

Yes |2x-5|>=4x+7|

Take the absolute value of both sides of the squared hail digging is simple.

3x+1)(x+6)<=0

The opening is upward, between the two, so x is on the interval [-6, -1 3].

xy|-2|x|+|y|=4

Turning the blind into :(|.)x|+1)(|y|-2)=2 because of the trouble|x|+1>=1 and is a simple bend space factor of 2.

So there is: x|+1=1,2

y|-2=2,1

i.e. |x|=0,1

y|=4,3

So there are 6 sets of integers.