If the square + X-1 = 0 of the equation AX about X has a real root, then the value range of a is $(acontent)

If the square + X-1 = 0 of the equation AX about X has a real root, then the value range of a is $(acontent)

On the square + X-1 = 0 of the quadratic equation AX with one variable of X has real roots
⊿=1²+4a≥0
4a≥-1
a≥-1/4
And a ≠ 0
The range of a ≠ a ≠ - 4
A=1
Then X-1 = 0
There are real roots
a≠0，
If there are real roots, then △ > = 0
1+4a>=0
a> = - 1 / 4 and a ≠ 0
To sum up
a>=-1/4
If a = 0, the equation is X-1 = 0, and there is a real root, the condition is satisfied when a = 0;
When a ≠ 0, the equation has a real root, that is, 1 + 4A ≥ 0, and the solution is a ≥ - 1 / 4
In conclusion, a ≥ - 1 / 4
b^2-4ac＞=0.
1^2-4a*(-1)>=0
That is 1 + 4A > = 0
a>=-1/4
Don't blame me for my mistake
Let real numbers a and B have real roots of equation X4 + AX3 + BX & # 178; + ax + 1 = 0, and find the minimum value of a & # 178; + B & # 178
But the answer is 4 / 5
Is it the fourth power of X
Given that the equation x ^ 4 + ax ^ 3 + BX ^ 2 + ax + 1 = 0 has real roots (a, B are real numbers), find the minimum value of a ^ 2 + B ^ 2
X ^ 2 + ax + B + ax ^ (- 1) + x ^ (- 2) = 0 (x + 1 / x) ^ 2-2 + a (x + 1 / x) + B = 0, let (x + 1 / x) = y, then y ^ 2 + ay + B-2 = 0 and X + 1 / x > 2 or = 2 or Y2 or y2 = Y2 ^ 2 + 4-9y2 ^ 2 / (Y2 ^ 2 + 1) = Y2 ^ 2 + 1 + 9 / (Y2 ^ 2 + 1) - 6. Because Y2 ^ 2 + 1 > = 5, the original formula > = 5 + 9 / 5-6 = 4 / 5
(x+1)^4=x^4+4x^3+6x^2+4x+1,
When x = - 1, x ^ 4 + 4x ^ 3 + 6x ^ 2 + 4x + 1 = 0, and x ^ 4 + ax ^ 3 + BX ^ 2 + ax + 1 = 0
x^4+4x^3+6x^2+4x+1=X^4+ax^3+bx^2+ax+1=0
When x = - 1, (x + 1) ^ 4 is the smallest, then a = 4, B = 6
So, the minimum value of a ^ 2 + B ^ 2 = 52
What is the range of the function y = - 2x + 3 (- 2 less than or equal to x less than 4)?
-2
Function y = - 2x + 3 (- 2 less than or equal to x less than 4)
When x = - 2, y = - 2 * - 2 + 3 = 7 is the largest
When x = 4, y = - 2 * 4 + 3 = - 5
So the range is (- 5,7]
Hope to help you
When x = - 2
y=(-2×-2)+3
=7
When X-2 × 4 + 3 = - 5
So his range is - 5
Finding sin2x and sin ^ 2 + sinxcosx with 2sinx cosx = 0
2sinx-cosx=0 2sin x=cos x tan x=sin x / cos x =1/2sin2x =2 sin x cos x =(2 sin x cos x ) / (sin^2 x + cos^2 x )=2tan x /(tan^2 x +1) =4/5sin^2 x +sin x cos x =(sin^2 x +sin x cos x) / (sin^2 x + cos^2...
Given that the function f (x) = 2x / X & # 178; + 6 is constant for any x > 0, f (x) t, the value range of t can be obtained
A:
F (x) = 2x / (X & # 178; + 6) for any x > 0, f (x) = 2 √ (x * 6 / x) = 2 √ 6 (properties of basic inequality or check function)
So:
Zero
tanx=4/3,π∈(π,π/2),sin2x=?cos2x=?tan2x=?
sinx/cosx=tanx=4/3
sinx=4/3*cosx
sin²x+cos²x=1
So cos & # 178; X = 9 / 25
sin2x
=2sinxcosx
=2(4/3*cosx)cosx
=8/3cos²x
=24/25
cos2x=2cos²x-1=-7/25
tan2x=2tanx/(1-tan²x)=-24/7
Function f (x) = {x + 2 (x ≤ - 1) / X & # 178; (- 1)
∵-3/2＜-1
∴f(-3/2)=x+2=-3/2+2=1/2
The second quadrant is costan = - 2x3 and is known
Ninety
Let f (x) = - X & # 178; + 2x = a (0 ≤ x ≤ 3, a ≠ 0) be m
Let f (x) = - X & # 178; + 2x = a (0 ≤ x ≤ 3, a ≠ 0) have the maximum value of M and the minimum value of n
(1) Finding the value of M and n (expressed by a)
(2) If the terminal edge of angle θ passes through point P (m-1, N + 3), the value of sin θ + cos θ + Tan θ is obtained
1) Axis of symmetry x = - 1
Monotonically increasing f (x) at 0 ≤ x ≤ 3
When x = 0, n = f (0) = 0
When x = 3, M = f (3) = - 3
(2) It is known from (1) that the coordinates of point P are (- 4,3)
sinθ=3/5,cosθ=-4/5,tanθ=-3/4
Original formula = (3 / 5) + (- 4 / 5) + (- 3 / 4)
=-11/20