If the equation x square 4x m has a real root, then the sum of all real numbers can be ?

Function y=|x²+4x|=|(x+2)²-4|images.

is to place the function y=(x+2) -4 above the x-axis.

Part of the image is preserved, with a portion below the x-axis.

The image is flipped up along the x-axis.

y=|x²+4x|=|(x+2)²-4|images.

About the straight line x=-2 symmetry.

Equation|x square + 4x|=mThe number of roots is.

y=|(x+2)²-4|The number of points where the image intersects with the line y=m.

When m=0, there are 2 intersections, the equation has 2 roots, and there is symmetry with respect to x=-2.

x1+x2=-4

At 04, there are 2 intersections, the equation has 2 roots, and x1=x2=-4 chooses d

1. When x 2 + 4x = m, that is, x 2 + 4x-m = 0, according to the discriminant formula of the root, 4 2-4*1*(-m) 0, that is, m -2, and because of |x square + 4x|=m, so m 0

2. When x 2+4x=-m, that is, x 2+4x+m=0, according to the discriminant formula of the root, 4 2-4*1*(m) 0, that is, 2 m, and because of |x square + 4x|=m, so 2 m 0

Based on the above, m cannot be negative, so none of the above answers are correct.

b From the meaning of the title, it can be seen that x square ten 4x=m and one x square one 4x=m, the two equations are subtracted to obtain 2x square ten 8x=0, 2 to get x square ten 4x=0, and then the two sides of ten 4 get x square ten 4x ten 4 = 4, then (x ten 2) square = 4,, x ten 2 = 2, x = 0 or x = a 4,, 0 ten (one 4) = a 4

If b squared knows that -4ac>=0, there will be a real root, and the corresponding parameter in the equation will be substituted:

b square - 4ac> = (2m) square - 4 * 1 * 4 (m-1) 4m square - 16m + 16

2m-4) square >=0

Therefore, the equation must have a real root.

x²-(m+2)x+4=0

If the equation has a real root, then the discriminant is greater than or equal to 0

(m+2)²-16≥0

m²+4x+4-16≥0

m²+4x-12≥0

m+6)(m-2)≥0

m -6 or m 2

Equation x squared -(m+2)x + 4 = 0 has a real root, then the discriminant of the root [-(m+2)] 4 4 0 then (m+2) -16 0

then (m+2+4)(m+2-4) 0

then (m+6)(m-2) 0

So m -6 or m 2

x²-(m+2)x+4=0

If the equation has a real root, then the discriminant is greater than or equal to 0

m -6 or m 2

m²+1)x²-2mx+(m²+4)=0

-2m) -4 (m +1) (m +4) 4m -4m 4-16m -4m -16-4 (m 4 + 4m +4).

4(m²+2)²

m²>=0

m +2>=2,4(m mulder closed +2) >84(m +2) <8, i.e. <0, so the equation about x (m squared + 1) x squared - 2mx + (m squared + 4) = 0 has no real cracks.

m Brigade pretending to be 1, m is not = 5. From the question, it can be obtained that ( x 2) 2=m 1 0 (if you take the equal sign, there are two roots that do not fit the topic), so m splits the answer 1, x =2 (m 1), so x= 2 is missing (m 1), when m = 5, x only has 3 solutions that do not fit the topic, so m belongs to (1,5)(5,+ solution.

It's so messy, can't you write it clearly?

x^2 - 4 | x|+5 =m i.e. ( x| -2)^2 = m-1

1.Four unequal real roots:

m-1>0, |x| -2 = m-1）， x|= 2 ±√m-1）x1 = 2 +√m-1）, x2 = 2 -√m-1）, x3 = 2 +√m-1）, x4 = 2 -√m-1）

by x2 ≠ x3 =>m≠5

When the M-1 > is delayed by 0 and m≠5, there are four real roots of unequal rock belts.

2.When m = 5, there are three unequal real roots.

Set x 2-4|x|+(5-m)=0, so that t=|x|It can be inferred that the bundle or: x 2-4t+(5-m)=0;

16-4*1*(5-m)>0;

4>5-m;m>1

At this time, t has two unequal values, in order to make each t rotten

x|After solving, there are two values, and a total of four values are disturbed, then.

t1，t2>0

So t1*t2=5-m>0, and t1+t2=4>0 (constant holds) [Veda's theorem].

In summary:1

There are 4 unequal real roots.

x|²-4|x|+3-m=0

That is, in the equation |x|There are two unequal positive roots.

So the discriminant formula is greater than 0

16-12+4m>0

m>-1

The positive root is multiplied by greater than 0

By the Vedic theorem.

i.e. 3-m>0

m<3 so -1

1) The domain of f(x)=3x-2 is r

2)f(x)=3|x|-2 is defined in the domain r

3) The domain of f(x)=3(x 2)+x-2 is r

4) f(x)=(3x-2) 0 is defined in the domain of 3x-2≠0, x≠2 3

So is (- 2 3) (2 3,+.)

5) f(x)=root number(3x-2) is defined in the domain of 3x-2>=0,x>=2 3

So it is [2 3,+

6) f(x)=5th root (3x-2) is defined in the domain r

7) f(x)=1 (3x-2) is defined in the domain of 3x-2≠0

So is (- 2 3) (2 3,+.)

8) f(x)=2x (3(x 2)-2) is defined as x ≠2 3, x ≠ 6 3

So it is (- 6 3) (6 3, 6 3) (6 3,+.)

9) The domain of the function y=3 (1-root number (1-x)) is

1-x>=0

x<=1

1-√(1-x)≠0

1-x≠1x≠0 so is (- 0) (0,1].

First of all, delta=4 2-4(3-m)=4(1+m)>0 m>-1 When x>=0, the equation is x2-4x+3-m=0, and there must be two non-negative roots, so x1+x2=4> difference 0, x1x2=3-m>=0 hall macro m<=3

When x<0, the equation is x2+4m+3-m=0, and there must be two negative roots, so x1+x2=-4<0, x1x2=3-m>0 m<3

So the sum gets: -1

Is x*x x x 2 pull? If yes, it's good to do...

Let y=x 2-4|x|+3 require, y=m

y=x^2-4|x|+3 is a piecewise function y={x 2-4x+3 x》0x 2+4x+3 x<0 This image can be drawn by Kiri Hao.

y=m has 4 intersections with this function trap (i.e. 4 intersections with the function image), then -1

There is no real root.

<0 ∴b²-4ac＜0

2（m+1））²4×（2m²+4）×1<04m²+8m+4-8m²-16<0

4m²+8m-12<0

4（m²-2m+3）<0

4(m²-2m+1）-8<0

4（m-1）²-8<0

m-1）²>=0

4（m-1）²<=0

4（m-1）²-8<0

Regardless of the value of m, the original equation does not have a real root.

That's pretty much it, detailed enough! Very powerful.

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