Discriminant of the root of quadratic equation with one variable We know the equation x & # 178; + (M + 2) x + 2m-1 = 0 (1) Proof: the equation has two unequal real roots (2) When m is the value, the two parts of the equation are opposite to each other, and the solution of the equation is obtained

1)delta=(m+2)^2-4(2m-1)=m^2-4m+7=(m-2)^2+3>0
So the equation must have two unequal real roots
2) If two are opposite numbers, then the sum of two = - (M + 2) = 0, that is, M = - 2
The equation is: x ^ 2-5 = 0
The roots are: √ 5 and - √ 5

On the equation (A-1) x ^ 2-2ax + a = 0 of X, when what is the value of a, the equation is a quadratic equation with one variable? When what is the value of a, the equation is a quadratic equation with one variable?

The coefficient of quadratic term is not zero, that is to say, when A-1 ≠ 0, a ≠ 1, the equation is quadratic equation with one variable
When the coefficient of quadratic term is zero and the coefficient of primary term is not zero, i.e. A-1 = 0, - 2A ≠ 0, i.e. a = 1, the equation is one variable linear equation

Discriminant of the root of quadratic equation of one variable (2) solution format
I forgot to take my notebook home. I didn't touch my math for a few days on national day
For example: when m takes what value, the equation 2x & # 178; - (4m + 1) x + 2m & # 178; - 1 = 0 (PS: there is no multiplication sign in the equation ~)
1) Have two unequal real roots?
2) Have two equal real roots?
3) No real roots?
Must be complete. The main problem is the format~
Thank you first

1、
Quadratic equation of one variable
ax²+bx+c=0
The discriminant is
△=b²-4ac
2、
When △ > 0, there are two unequal real roots;
When △ = 0, there are two equal real roots;
△

It is known that the quadratic equation of one variable with respect to X is KX + (2k-3) x + (K-3) = 0

x=1、 x=(k-3)/k

It is known that the quadratic equation of one variable x ^ 2-4kx + 1 / 4 = 0 (k is a constant greater than zero) has two equal real roots, then K=_ The two equations are equal
The real root of is_

From the question, △ = (4K) & # 178; - 4 × (1 / 4) = 16K & # 178; - 1 = 0
The solution is k = ± 1 / 4
The original equation is X & # 178; ± x + 1 / 4 = (x ± 1 / 2) &# 178; = 0
The solution is x = ± 1 / 2
So, if the equation has two equal real roots
Then, k = 1 / 4, and the two equal real roots of the equation are x = 1 / 2
Or, k = - 1 / 4, the two equal real roots of the equation are x = - 1 / 2

If - 2 is a root of the quadratic equation (k2-1) x2 + 2kx + 4 = 0 with respect to x, then K=______ ．

∵ - 2 is a root of the univariate quadratic equation (k2-1) x2 + 2kx + 4 = 0 with respect to x, and the solution is k = 0 or 1. When k = 1, the equation is not a univariate quadratic equation, so k = 0

I have the answer to the problem of filling in the blanks,
1. If the equation x2 + 2x-1 + M = 0 has two equal real roots, then M=__ ．
2. A is a rational number, B is a rational number The root of the equation 2x2 + (a + 1) x - (3a2-4a + b) = 0 is also rational
3. When k ＜ 1, equation 2 (K + 1) x2 + 4kx + 2k-1 = 0 has The root of a real number
5. If the quadratic equation MX2 + 3x-4 = 0 with respect to X has real roots, then the value of M is ．
6. If the equation 4mx2 MX + 1 = 0 has two equal real roots, then M is ．
7. In the equation x2 MX + n = 0, m and N are rational numbers, and one root of the equation is 2
8. In the quadratic equation AX2 + BX + C = 0 (a ≠ 0), if a, B, C are rational numbers and Δ = b2-4ac is a complete square number, then the equation must have ．
9. If M is a nonnegative integer and the quadratic equation (1-m2) x2 + 2 (1-m) X-1 = 0 has two real roots, then the value of M is ．
10. If the quadratic equation kx2 + 1 = x-x2 about X has real roots, then the value range of K is ．
11. Given that the equation 2x2 - (3m + n) x + M &; n = 0 has two unequal real roots, then the value range of M and N is ．
12. If the two real roots of the equation a (1-x2) + 2bx + C (1 + x2) = 0 are equal, then the relation of a, B, C is_ ．
13. If the quadratic equation (k2-1) x2-6 (3K-1) x + 72 = 0 has two real roots, then K is_ ．
14. If the quadratic equation (1-3k) x2 + 4x-2 = 0 has real roots, then the value range of K is ．
15. The case of equation (x2 + 3x) 2 + 9 (x2 + 3x) + 44 = 0 is solution
16. If the equation x2 + PX + q = 0 has equal real roots, then the equation x2-p (1 + Q) x + Q3 + 2q2 + q = 0 Real roots
1．2
2．1
3. There are two unequal
4．6,-4
5. M ≥ - 9 / 16 and m ≠ 0
6．16
7．4,1
8. Two rational roots
9．m=0
K ≤ - 3 / 4 and K ≠ - 1
M, n is any real number not equal to zero
12．b2-c2+a2=0
13. Any real number
14．k≤1
15. No real number
16. There are also equal
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1. In this equation, a = 1, B = 2, C = - 1 + m, so the square of discriminant = 2 - 4 × 1 × (- 1 + m) = 8-4m. Because the equation has two equal real roots, the discriminant = 8-4m = 0, so m = 2.3. Discriminant = 16K square - 8 × (K + 1) × (2k-1) = - 8K + 8, because K ＜ 1, so - 8K ＞ - 8, - 8K + 8 ＞ 0, so

Two univariate quadratic equations with (radical 5) - 4, (radical 5) + 4 and coefficient of first order 2

Let ax ^ 2 + 2x + B = 0
Then X1 + x2 = 2 times (radical 5) = - 2 / A,
x1*x2=-11=b/a
The solution is a = - 1 / (radical 5),
B = 11 / (radical 5)
The equation is [- 1 / (radical 5)] x ^ 2 + 2x + [11 / (radical 5)] = 0

According to the following requirements, the quadratic equation of one variable with quadratic coefficient 1 can be solved. (1) two are 2 + radical 3 and 2-radical 3 respectively
(2) The sum of the two is 3 and the product is - 4
2. For the equation x2-2 (K + 1) x + K2 = 0 of X, both of them are positive numbers to find the range of K
3. Given that the sum of the two reciprocal of the equation x2-2x + k = 0 is 8 / 3, find the value of K

How can we get the 2k-3k-2 = 0 from this | 3K-1 | = root sign (5 + 5K Square)

|3K-1 | = radical (5 + 5K)
Two sides square
（3k-1）²=5+5k²
9k²-6k+1=5+5k²
4k²-6k-4=0
Divide both sides by 2:2k-3k-2 = 0

Discriminant of the root of quadratic equation with one variable We know the equation x & # 178; + (M + 2) x + 2m-1 = 0 (1) Proof: the equation has two unequal real roots (2) When m is the value, the two parts of the equation are opposite to each other, and the solution of the equation is obtained