Given that X1 and X2 are the two real roots of the equation x ^ 2-2x-2 = 0, we can solve the equation to find the value of x2 + (2 △ x1)

Given that X1 and X2 are the two real roots of the equation x ^ 2-2x-2 = 0, we can solve the equation to find the value of x2 + (2 △ x1)

Hello, Lou Lou
This problem needs to use Veda's theorem
x²-2x-2=0
According to Weida's theorem, X1 + x2 = 2, x1x2 = - 2
And the title requires x2 + 2 / X1 = (x1x2 + 2) / X1 = (- 2 + 2) / X1 = 0
If you have any questions, please ask

It is known that the equation (k-1) × x (2k-3) x + K + 1 = 0 has two unequal real roots x1, x2
1. Find the value range of K;
2. Whether there is a real number k is that the two real roots of the equation are opposite to each other? If there is, find out the value of K; if not, explain the reason

1. It can be concluded that if k-1 ≠ 0, then K ≠ 1
△=（2k-3）²-4（k-1）（k+1）=4k²-12k+9-4k²+4= -12k+13＞0
Then K < 13 / 12 and K ≠ 1
2. According to Weida's theorem:
x1+x2= -（2k-3）/（k-1）=0
Then: - (2k-3) = k-1
3k=2
k=2/3

It is known that the equation x ^ 2 + (2k-3) x + K ^ 2 + 1 = 0 has two unequal real roots x1, X2, and x1, X2 are greater than 1
Ask if X1 / x2 = 1 / 2 to find the value of K

∵ has two unequal real roots, and both X1 and X2 are greater than 1
∴(2k-3)^2-4(k^2+1)>0,-(2k-3)/2>1
∵x1/x2=1/2
∴x2=2x1
∵x1+x2=-(2k-3)
∴x1=(3-2k)/3
∵x1*x2=k^2+1
∴ x1^2=(k^2+1)/2
∵ x1, X2 are all greater than 1
∴(3-2k)/3>1
(k^2+1)/2>1
∴ k

It is known that the equation k2x2 + (2k-1) x + 1 = 0 has two unequal real roots x1, x2
(1) Find the range of K;
(2) Is there a real number k so that the two real roots of the equation are opposite to each other? If so, find out the value of K. if not, explain the reason
I got the answer. It's K

(1)
The equation has two unequal real roots
So the solution of △ > 0 is K0
So K

It is known that: for the equation x2-kx-2 = 0. (1) proof: no matter what the value of K is, the equation has two unequal real roots. (2) let the two roots of the equation be x1, X2, if 2 (x1 + x2) > x1x2, find the value range of K

(1) It is proved that: from the equation x2-kx-2 = 0, we know that a = 1, B = - K, C = - 2, ∧ = b2-4ac = (- K) 2-4 × 1 × (- 2) = K2 + 8 > 0, ∧ no matter what the value of K is, the equation has two unequal real roots; (2) ∫ the two roots of equation x2-kx-2 = 0 are x1, X2, ∧ X1 + x2 = k, x1x2 = - 2, and ∫ 2 (x1 + x2) > x1x2, ∧ 2K > - 2, that is, K ＞ - 1

If the equation | x2-2x | + m + 1 = 0 of X has two unequal real roots, then the value range of M is______ ．

The equation | x2-2x | + m + 1 = 0 of X has two unequal real roots, that is, the function y = | x2-2x | and y = - M-1 have two different intersections. From the graph, when y = - M-1 coincides with X axis or above y = 1, it is consistent, that is - M-1 = 0 or - M-1 > 1 & nbsp; the solution is m = - 1 or m < - 2, so the answer is: M = - 1 or m < - 2

If the equation x ^ 2 - 2x + M = 0 has two unequal real roots, then the value range of is?

X^2 - 2x + m = 0
(-2)^2-4*1*m>0
4-4m>0
m

If the equation MX2 - (m-1) x + 1 = 0 has Reagan, find the value of M

Hello
The original formula is (MX-1) (x-1) = 0
△=b2-4ac
=(m-1) 2-4m = > m2-6m + 1 is the square of a rational number
When m = 0 or M = 6, m2-6m + 1 = 1
So m = 0 or M = 6
The above is for reference only

If the equation f (x) = MX2 + 2 (M + 1) x + m + 3 = 0 has at least one negative root, then the value range of M is______ ．

When m = 0, the equation is 2x + 3 = 0, and has a negative root. When m ≠ 0, MX2 + 2 (M + 1) x + m + 3 = 0 is a quadratic equation of one variable. If there are 0 roots, then M + 3 = 0,  M = - 3, and the equation is - 3x2-4x = 0, and has a negative root. Suppose that the equation has no one negative root and 0 root, then the equation has no real root or two positive roots, and let the roots be x1

It is known that X1 = - 1 is a root of the equation x2 + MX-5 = 0. Find the value of M and the other root of the equation x2

From the meaning of the question: (- 1) 2 + (- 1) × m-5 = 0, the solution is m = - 4; when m = - 4, the equation is x2-4x-5 = 0, the solution is X1 = - 1, X2 = 5, so the other root of the equation is x2 = 5