It is known that the quadratic equation x2-2x-m = 0 with respect to X has real roots. (1) find the value range of M; (2) if a and B are the two roots of the equation and satisfy (12a2 − a + 1) (2B2 − 4B − 1) = 32, find the value of M

It is known that the quadratic equation x2-2x-m = 0 with respect to X has real roots. (1) find the value range of M; (2) if a and B are the two roots of the equation and satisfy (12a2 − a + 1) (2B2 − 4B − 1) = 32, find the value of M

(1) ∵ x2-2x-m = 0 has a real root, ∵ a2-2a = m, b2-2b = m, and (12a2-a + 1) (2b2-4b-1) = 32, ∵ (12m + 1) (2m-1) = 32, that is, (2m + 5) (m-1) = 0, then 2m + 5 = 0 or M-1 = 0, and M = 1 or M = - 52 (rounding off)

Write a necessary and sufficient condition for the quadratic equation x ^ 2-2x + m ^ 2 = 0 with real roots

2^2 - 4m^2 = 4(1 - m^2) >= 0
-1

Given two equal real roots of the quadratic equation x2 + 2x-m = 0 with respect to x, then the value of M is?
Change this x2 + 2x-m = 0 to X & # 178; + 2x = 0

∵ the equation x & # 178; + 2x-m = 0 has two equal real roots
The discriminant △ = 2 & # 178; - 4 × 1 × (- M) = 4 + 2m = 0
∴4m=-4
∴m=-1
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Using factorization to solve quadratic equation of one variable 2x + 7x + 6 = 0 (2x -?) (x -?) = 0
Ax ^ 2 + BX + C = 0, a = 2, B = 7, C = 6, there is no other constant except a, B, C
(2x -?) (x -?) = 0 what are the two? What are the two? Ax ^ 2 + BX + C = 0? How many letters are there in total? Are there any paradoxes?
Example:
4/3x^2+14/3x+4=0
（4/3x+√4)(x+√4)=0
x1=-3/2 x2=-2

-3 and - 2,

Univariate quadratic equation 2x2-2 √ 2x + 1 = 0
The answer is X1 = x2 = √ 2 / 2
I just want to ask if this problem can be solved by factorization and find out the reason

2X ^ 2-2 √ 2x + 1 = 0, 2 * x ^ 2 can be regarded as (√ 2x) ^ 2. In this way, the equation can be simplified to T ^ 2-2t + 1 = 0 (t = √ 2x), which is the complete square formula. It is easier to solve t = 1 = √ 2x, so X1 = x2 = √ 2 / 2. As for whether you can use factorization, it is OK, but the requirement is relatively high