It is known that the equation (k-1) x ^ 2 + (2k-3) K + K + 1 = 0 has two unequal real roots a and B Is there a real number k so that the two real numbers and roots of the equation are opposite to each other? If so, find out the value of K; if not, explain the reason

It is known that the equation (k-1) x ^ 2 + (2k-3) K + K + 1 = 0 has two unequal real roots a and B Is there a real number k so that the two real numbers and roots of the equation are opposite to each other? If so, find out the value of K; if not, explain the reason

If the equation has two unequal real roots, then it is more than 0
(2K-3)²-4（K+1)(K-1）>0
K

If K is a real number, then there are two unequal roots of the equation x ^ 2 + (2k + 1) x + k-1 = 0

The discriminant Δ = (2k + 1) &# 178; - 4 (k-1) = 4K & # 178; + 4K + 1-4k + 4 = 4K & # 178; + 5 > 0
So the above equation has two unequal real roots

On the equation x ^ 2-px + 1 = 0 of X (o belongs to two roots x1, X2 of R), and | x1 | + | x2 | = 3, the value of P is obtained

According to Weida's theorem: X1 + x2 = P, X1 * x2 = 1, so X1 and X2 have the same sign
|X1 | + | x2 | = 3 when both X1 and X2 are less than 0, - x1-x2 = 3, X1 + x2 = P = - 3
When both X1 and X2 are greater than 0, X1 + x2 = 3, X1 + x2 = P = 3

Given the two roots x 1 and x 2 of the equation x ^ 2-P x + 1 = 0 (P belongs to R), and x 1 + x 2 = 3, find the value of P
Why X1 and X2 are not imaginary numbers? How to prove that P belongs to (- infinity, - 2) U (2, + infinity)

If the discriminant = P ^ 2-4 > = 0, so p ^ 2 > = 4, so p = 2, according to the relationship between the root of quadratic equation and coefficient, X1 + x2 = P, X1 * x2 = 1, because X1 * x2 = 1 > 0, so X1 and X2 have the same sign, when X1 and X2 are positive numbers, 3 = | x1 | + | x2 | = X1 + x2 = P, so p = 3; when X1 and X2 are negative numbers, 3 = | x1 |

There are several ways to divide the area of a parallelogram into two parts

There are four kinds of straight lines, and countless kinds of broken lines, curves and so on

There are several ways to fold a piece of paper in a parallelogram so that the cross fold of the paper bisects half of the area of the parallelogram
a.1
b.2
c.3
d. Countless

Countless, as long as through the intersection of parallelogram diagonal can, such as L1 to L5 are

Can any parallelogram be cut into rectangle by cutting method?
Can we cut any parallelogram into rectangle by cutting?
Note: is "arbitrary", this parallelogram includes rectangle, square, diamond, parallelogram
It seems impossible to cut a square, but the students said they could cut it into small pieces like a cake and then put it together
┌┬┬┬┐
├┼┼┼┤
└┴┴┴┘

Yes,
4 the closest two points in the endpoint
Make a vertical line to the opposite side respectively!

Why can every parallelogram be converted into a rectangle

A rectangle itself is a parallelogram. To convert it, it is also a parallelogram without a right angle in the figure. The method is to cut it obliquely in the middle and put two parts with right angles together

Why cut parallelogram into rectangle along the height of parallelogram

Because along the high shear can be cut into a right angle

A parallelogram is drawn into a rectangle
I want to flash a parallelogram into a rectangle

Make deformation animation is OK ~ I'm charging, do not know you can accept it? If you can add 501357543