Junior 3 Mathematics, Simple, Junior High School Simple Mathematics

By the title: Let equation 1The other root is x, x*0=m2-2m-3=0m=-1 or 3

When m = -1 and x 2 = 0, there are 2 equal real roots, so they are discarded.

So m=3

Equation 2Reduced to: x 2-(k-3)x-k-22=0x1+x2=k-3

x1x2=-k-22

x1-x2|=root(x1-x2) 2=root[(x1+x2) 2-4x1x2].

The root number [(k-3) 2+4(k+22)]=1, i.e., k2-2k+96=0

Because the discriminant formula: (-2) 2-4*96<0

So there is no solution to the equation, so there is no value of k.

Give me the score, right? Hey! I don't understand hi me.

Solution: Substituting one of the two unequal real roots of the known condition " is 0 "to obtain:

m -2m-3 = 0, the solution is m = -1 or m = 3, considering that when m = -1, the two roots are 0, which is inconsistent with the condition that "one of the two unequal real roots is 0", so it is rounded, so m = 3

It can be simplified to x - (k-3) x - k-22 = 0

From the relation between the root and the coefficient, we get: x1+x2=(k-3), x1x2=-(k+22) thus, |x1-x2|x1-x2) x1+x2) -4x1x2] [k-3) +4(k+22)] k-1) +96] 96 Therefore, there is no k value, such that the absolute value of the difference between x1 and x2 is 1

Solution: Substituting x 0 into the equation

Solution: m 3 m 1

When m 3 , the equation becomes: x (k 3) x k 22 0 from Veda's theorem: x1 x2 k 3 (a) x1·x2 k 22 (ii).

Obtained from (1) 4(2): k 6k 9 4k 88 (x1 x2) 1

k²－2k＋96＝0

Find: 0 No real number solution.

When m 1 and the equation is x 0 there are two identical solutions to the real numbers, which does not fit the problem.

To sum up, it does not exist.

According to the title, the value of the discriminant of the root needs to be a perfect square number: 2 2 (m + 1) 2 - 4m 2 = 4 (m 2 + 2m + 1 - m 2) = 4 (2m + 1) is a perfect square number.

Therefore (2m + 1) must be a perfectly square number.

And there is [2(m + 1) +2 (2m + 1) ] 2= m + 1 + 2m + 1) 40m + 2m + 1) 39

m ＜ 39

According to the scope there are:

25 2m + 1 79, i.e. find the exact square number between 26 and 79 (and be an odd number).

There is a solution of 2m + 1 = 49 and m = 24

All integers between 13 and 39 are acceptable.

It is a general form that requires the equation to be reduced to a unary quadratic equation ax +bx+c=0(x+m) 2+n=0

x²+2mx+ m²+n=0

So the result is: x +2mx + m +n=0

It is to find the solution of the equation, that is, to find the value of x.

Take (x+m) and use the full square formula to get x square + m square + 2xm

Solution: Angle acb=90 degrees ac=3 ab=5 bc=4

Let the height of the AB side be h

Easy to get h=12 5

The RT triangle rotates once on the axis of the line where the AB side is located: S side = 12 5 3 12 5 4 = 84 5

The rotation formed two cones, the bottom surfaces of the two cones coincide, the radius is equal, equal to the height on the AB edge, and the radius of the sector on their sides is the length of AC and BC respectively.

The radius of the base circle of the cone is:

bc=√(5²-3²)=4

1 2 3 4 = 1 2 5 ab high.

The height of ab = the radius of the base circle.

The side area of the sector with AC radius is:

The side area of the sector with BC radius is:

Total side area =

The side area of the cone is equal to the area of the semicircle, and the figure is a semicircle with a radius of 12cm.

Side area = 1 2 12 12 = 72 (cm).

1 Yes, in mathematics there are two conditions for drawing a definite circle with a ruler.

The first is: the center of the circle.

The second is: a definite radius.

Draw a circle with the diameter of the known line segment AB", the center of the circle is the midpoint of the line segment AB, and the radius is half of the length of the line segment AB, and only the only circle can be drawn.

Draw a circle with the length of the known line segment AB as the diameter", the radius is determined, which is one-half of the line segment AB, but the center of the circle is uncertain, so this condition cannot make a unique circle, and an infinite number of circles can be drawn with such a condition.

So, the answer to this question: the first condition can make a unique circle, and the second can make an infinite number of them, which are not the same.

2 Strictly speaking, a wheel cannot be considered a circle.

Geometry defines that when a line segment rotates in a plane around one of its endpoints, the trajectory of its other end is called a circle.

It is true that you said that the outermost line of the wheel can be regarded as a circle, but the wheel is a surface, not a trajectory, so the wheel is a circle and not a circle.

In fact, there is a potential definition of geometry: the movement of points into lines, the movement of points can form a line (for example, drawing a straight line with a pencil), and the movement of lines into a surface (for example, the rotation of a bicycle strip to form a circular surface), and mastering this point will greatly help you understand space.

The circle here is a surface, not a line, the circle is determined by the center of the circle and the radius, the center of the circle determines the position of the circle, the radius determines the size of the circle, and the different centers of the circle, although the radius is equal, cannot be said to be the same circle.

(1) Solution: If the problem is 5k, then AE is 3k, and dc is 8k. Folded EF=BE=5K

In the RT triangle AEF, the Pythagorean theorem is obtained: AF=4K proves AEF DFC

ae：df=af：dc 3k：df=4k：8k∴df=6k

bc=ad=af+df=10k

In RT EBC, the Pythagorean theorem is obtained: EC 2 = Be 2 + BC 2 = (5K) 2 + (10K) 2 = (15 5) 2

k=3, so ab=8k=24 bc=10k=30(2) let ab cut circle o to m cut bc to n with om, on then bm = on=om and emo is similar to ebc

om:bc=em:eb=(eb-mb):eb i.e. r:30=(15-r):15 r=10

Circle area = 100

m3-2m-1=m3+1-2(m+1)=(m+1)(m2-m+1)-2(m+1)=(m+1)(m2-m-1)

If m is the root of x2-x-1=0, then m2-m-1=0, and Acer makes the edge pants m3-2m-1=0, m3=2m+1