It is known that the univariate quadratic equation x ^ 2 - (m-2) x + 1 / 4m ^ 2-2 = 0 1) When m is a value, the equation has two equal real roots 2) If the two real roots X1 and X2 of this equation satisfy X1 ^ 2 + x2 ^ 2 = 18, find the value of M

It is known that the univariate quadratic equation x ^ 2 - (m-2) x + 1 / 4m ^ 2-2 = 0 1) When m is a value, the equation has two equal real roots 2) If the two real roots X1 and X2 of this equation satisfy X1 ^ 2 + x2 ^ 2 = 18, find the value of M

1.b^2-4ac=0
(m-2)^2-m^2+8=0
M=3
When m is 3, the equation has two equal real roots
2
x1+x2=m-2
x1x2=1/4m^2-2
x1^2+x2^2=(x1+x2)^2-2x1x2
=1/2m^2-4m+8
Because X1 ^ 2 + x2 ^ 2 = 18
So 1 / 2m ^ 2-4m + 8 = 18
M1 = 10, M2 = - 2
One ascend tower is equal to zero, two solve the first order equation of three variables
1. Discriminant = 0 m = 3
2. When X1 + x2 = (m-2) X1 * x2 = 1 / 4m ^ 2-2, M = 10 or M = - 2 is substituted into the test m = 10, the discriminant is less than 0, and is discarded. The answer is m = - 2
1.△=0
2. X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2, it's OK to set up Weida's theorem
Solution equation: 3Y ^ 2 - (2Y + 1) (Y-1) = (Y-5) (y + 1)
∵3y^2-(2y+1)(y-2)=(y-5)(y-1)
==>3y^2-2y^2+3y+2=y^2-6y+5
==>9y==3
==>y=1/3
The solution of the original equation is y = 1 / 3
I hope it can help you,
If the function f (x) = x & # 178; - ax + 5 is a monotone function on [- 1.2], then the value range of real number a
f(x)=x²-ax-5
With the opening upward, the axis of symmetry is x = A / 2, decreasing on the left and increasing on the right
So:
(1) A / 2 ≥ 2, a ≥ 4
(2) A / 2 ≤ - 1, a ≤ - 2
In conclusion, the value range of a is a ≤ - 2 or a ≥ 4
A / 2 ≥ 2 or a / 2 ≤ - 1
A ≥ 4 or a ≤ - 2
The function f (x) = x & # 178; - ax + 5 is a monotone function on [- 1.2], which shows that the axis of symmetry is not in this interval,
Therefore, the axis of symmetry x = A / 2 ≥ 2 or x = A / 2 ≤ - 1
Therefore, a ≥ 4 or a ≤ - 2
f(x)=x²-ax+5=（x-a/2）^2-a^2/4-5
So there are two cases.
1) If [- 1.2] interval belongs to the descending interval, then a / 2 > = 2, then a > = 4
2) If [- 1.2] interval belongs to ascending interval, then a / 2
First simplify, then evaluate: 2 (3x2 + y) - (2x2-y), where x = 12, y = - 1
When x = 12, y = - 1, the original formula = 4 × (12) 2 + 3 × (− 1) = 4 × 14 + (− 3) = 1 + (- 3) = - 2
We know the quadratic equation of one variable x square - 2x + M-1 about X
(1) When m takes what value, the equation has two unequal real roots?
(2) Let X1 and X2 be the two real roots of the equation, and satisfy X1 square + x1x2 = 1, find the value of M
The original equation is x square - 2x + M-1 = 0
X square - 2x + M-1 is just a formula. There is no equal sign. Is it "x square - 2x + M-1 = 0"?
2Y ^ 2-y-6 = 0
2y^2-y-6=0
(2y+3)(y-2)=0
Y = - 3 / 2 or y = 2
Given that the function f (x) = X3 + 2x2-ax + 1 has an extreme point in the interval (- 1,1), then the value range of real number a is______ .
If f ′ (x) = 3x2 + 4x-a, when f ′ (- 1) f ′ (1) < 0, the function f (x) = X3 + 2x2-ax + 1 has an extreme point in the interval (- 1, 1), and the solution is - 1 < a < 7. If a = - 1, f ′ (x) = 3x2 + 4x + 1 = 0, there is an X = - 13 in (- 1, 1). If a = 7, f ′ (x) = 3x2 + 4x-7 = 0, and there is no real root in (- 1, 1), then the value range of a is - 1 ≤ a < 7, so the answer is given It is - 1 ≤ a ＜ 7
Where (supy) = - 3x; - 4x; - 1
（3x-4y）²-(3x+4y)²-xy
=（3x-4y-3x-4y）（3x-4y+3x+4y）-xy
=（-8y）（6x）-xy
=-49xy
Bring in x = 1, y = - 1
The original formula is 49
It is known that the quadratic equation K (X & sup2; - 2x + 1) - 2x & sup2; + x = 0 with respect to X has two real roots
Because K (X & # 178; - 2x + 1) - 2x & # 178; + x = 0, it is sorted out as follows: KX & # 178; - 2kx + k-2x & # 178; + x = 0, namely: (K-2) x & # 178; - (2k-1) x + k = 0
Because the equation has two solutions, so K-2 ≠ 0, K ≠ 2;
And the discriminant △ = B & # 178; - 4ac = (2k-1) &# 178; - 4 (K-2) k ≥ 0 is sorted out as: 4K + 1 ≥ 0, so: K ≥ - 1 / 4 and K ≠ 2;
Two x-three y = 1 x + 2Y = 2 solution equation
x/2-y/3=1
Simplification: 3x-2y = 6
x+2y=2②
①+②
3x+x=6+2
4x=8
X=2
2+2y=2
2y=0
Y=0
That is: the solution of the equations is x = 2; y = 0
(1) 2 / x-3 / y = 1 -- the result is (3) 3x-2y = 6
（2）x+2y=2
(2) Add to (3)
4x=8
Substituting x = 2 into (2)
2y=0
Y=0