How to solve the absolute value equation 1 b 2 4 b 5 and why

There are two ways to do this:

The first is the discussion, the principle is "if the number in the absolute value is greater than or equal to 0, then its value is equal to itself, if it is less than 0, it is equal to its opposite number", for this problem, 1-b = 2, when 1-b 0, 1-b = 1-b, that is, 1-b=2, then b = -1, when 1-b<0, 1-b = -(1-b), that is, -(1-b)=2, then b= can try this method 4+b =5, (the answer is 1 or -9).

We all know that the square of b is equal to b 2, then the square of 1-b = (1-b) 2, that is, (1-b) 2 = 2, simplified to b 2-2b-3 = 0, the solution is b = -1 or 3, (not recommended).

1-b|=2 means 1-b=2 or 1-b=-2, i.e., b=-1 or b=3, in the same way, 4+b=5 means 4+b=5 or 4+b=-5, i.e., b=1 or b=9

Remove the absolute value sign and consider both plus and minus 2 cases.

1-b|=2 rule.

1 -b =2 or 1 - b = -2

b =-1 or b=3

4+b| =5

4+b =5 or 4+b = -5

b =1 or -9

For equations with absolute values, the absolute values are removed first, and then the equations are solved.

x²-2x-8|=40 can become:

x²-2x-8=40

Or. x²-2x-8=-40

Then solve them separately.

According to the meaning of the absolute value, it can be obtained:

x —2x—8 = 40, and then solve two unary quadratic equations separately.

|x²-2x-8|=40

i.e.: x -2x-8 = 40 or x -2x-8 = -40x -2x-8 = 40

x²-2x-48=0

x-8）（x+6）=0

x=8 or x=-6

2、x²-2x-8=-40

x²-2x+32=0

x-1）²+32=0

No solution. So x -2x-8|The solution of =40 is: x=8 or x=-6

Solution: 9 a 2+4 b 2=1 (1)a 2--b 2=5 (2) from (2).

a 2 = b 2 + 5 (3) Substitute (3) for (1) to get:

9/(b^2+5)+4/b^2=1

Multiply both sides by b 2 (b 2 + 5) to get:

9b^2+4(b^2+5)=b^2(b^2+5)9b^2+4b^2+20=b^4+5b^2b^4--8b^2--20=0

b^2--10)(b^2+2)=0

Because b 2+2>0

So b 2--10=0

b^2=10

So a 2=b 2+5

Lack of respect 2x-1 Brigade side -1 = 2

So 2x-1 -1= 2

2x-1│-1=2

2x-1│=3

2x-1= ±3

Solution. x=2 or x=-1

2x-1│-1= -2

2x-1│= 1

It is not dismantled and dressed.

In summary. x=2 or x=-1

Interval analysis, removing absolute values.

1) When m<0, the inequality is 1-m>-m, so 1>0 is constant, so the solution is m<0;

2) When 0<=m<1, the inequality is 1-m>m, so m<1 2, so the solution is 0<=m<1 2;

3) When m>=1, the inequality is -(1-m)>m, hence -1>0, and therefore the solution set is empty;

Taking the union of the above three, the solution set of the original inequality is {m | m<1/2｝。

Considered in 3 paragraphs.

If m<0, the inequality is: 1-m>-m 1>0 The solution is: m<0 If 0<=m<1, the inequality is: 1-m>m 2m<1 m<1 2 The solution is: 0<=m<1 2

If m>1, the inequality is: m-1>m -1>0 contradictory, no solution In short, the solution of the inequality is:- m<1 2.

││2x-1│-1│=2

So 2x-1 -1= 2

2x-1│-1=2

2x-1│=3

2x-1= ±3

The solution yields x=2 or x=-1

2x-1│-1= -2

2x-1│= -1

It is not true that the sum of x = 2 or x = -1 is not established

Because + -2 = 2

So 2x-1 -1 = + -2 so 2x-1 = 3 or -1 when 2x-1 = 32x-1= +3 or -3 so 2x = 4 or -2x = 2 or -1 when 2x-1 = -1, this is not possible, there will be no new solution.

So, the answer is 2 or -1.

Known (3b+2) +2c+1) = 0

Then 3b+2=0, 2c+1=0

So b = -2 3 and c = -1 2

To make the equation ax +bx+c=0 have two unequal real roots, then δ=b -4ac=4 9-4a*(-1 2)=4 9+2a 0, so a -2 9

If you don't understand, please ask, and I wish you a happy study!