How to calculate that x square plus ax minus a square equals 0

How to calculate that x square plus ax minus a square equals 0

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The process of solving the real roots of two roots where x square plus ax minus a equals zero
X ^ 2 + ax-a = 0 has two real roots
∴a^2+4a≥0
A ≥ 0 or a ≤ - 4
x^+ax-a=0
x^+ax+a^2/4-a-a^/4=0
(x+a/2)^2=(4a+a^)/4
X = - A / 2 + (-) radical (4a + A ^ 2) / 2
=[- A + - radical (4a + A ^ 2)] / 2
Given that f (x) = 2x + 1, X belongs to the domain of [1.9] finding f (x + 1) + F (x's Square)
Urgent ~ ~ ~ good bonus
My answers are (1,3) and (8,28), right
Because x ∈ [1,9], then 1 "x + 1" formula 9.1, 1 "X & sup2;" formula 9.2 is derived from Formula 1, 0 "X" formula 8,2 can be transformed into Formula 1,. 4, X & sup2; "formula 9,5 is derived from formula 4, or X-1, formula 5 is derived from formula 4-3" X "3, starting from formula 4 and formula 5, - 3 ≤ x ≤ - 1, or 1 ≤ x ≤ 3.6, then combining Formula 1 and formula 6
One
If the solution set of the inequality ax ^ 2 + BX + C > 0 (a is not equal to 0) is an empty set, then the following formula is correct
A.a0
B.a
There are also reasons, why is this the answer?
C.
Because if the equation AX ^ 2 + BX + C = 0 (a > 0) has no real number solution, that is, the solution set of B ^ 2-4ac0 is all real numbers, while the solution set of inequality ax ^ 2 + BX + C0 (a is not equal to 0) is empty
So a
BC
The solution set of ax ^ 2 + BX + C > 0 (a is not equal to 0) is an empty set
It shows that there is no point above the x-axis on the modified parabola
So open your mouth down
There is no intersection with X axis or only one intersection Δ ≤ 0
C a
The range of function y = - 3x + 1, X ∈ {- 1,1} is
X = 1, minimum = - 3 + 1 = - 2
X = - 1, max = 3 + 1 = 4
Range [- 2,4]
Just insert it
x=-1
Y=4
X=1
y=-2
Ψ range [- 2,4]
If the solution set of inequality ax square + BX + C is less than 0 (a is not equal to 0), then?
A. A is less than 0, △ 0, B. A is less than 0, △ less than or equal to C. A is more than 0, △ less than or equal to 0, d. A is more than 0, △ greater than or equal to 0
B
If the solution set of inequality ax square + BX + C is less than 0 (a is not equal to 0), then
A
The function y = 3x / (x * x + X + 1) where X "0", find the range
y=3x/(x^2+x+1) =3/（x+1/x+1）
By X
Y = 3x / (x * x + X + 1) = 3 / (x + 1 / x + 1)
If the solution set of inequality ax ^ 2 + BX + 1 > 0 is x not equal to 2, then the value of a B is
Then x = 2 is the unique solution of the equation AX ^ 2 + BX + 1 = 0
It is also the solution of x ^ 2 + BX / A + 1 / a = 0
From Veda's theorem, 2 * 2 = 1 / A, 2 + 2 = - B / A
A = 1 / 4; b = - 1
The method of discriminant to calculate the range of value
What form should we use this method? How to use it? What should we pay attention to?
For the fractional function y = f (x) = (AX ^ 2 + BX + C) / (DX ^ 2 + ex + F), the
Because for any real number y, its necessary and sufficient condition in the range of function f (x) is that the equation y = (AX ^ 2 + BX + C) / (DX ^ 2 + ex + F) about X has a real solution, so "finding the range of value of F (x)". This problem can be transformed into "given that the equation y = (AX ^ 2 + BX + C) / (DX ^ 2 + ex + F) about X has a real solution, finding the range of value of Y."
In this paper, we take x as an unknown quantity and y as a constant, and transform the original formula into a quadratic equation with one variable (*) about X, so that the equation has a real solution
(1) When the coefficient of quadratic term is 0, the corresponding value of Y is substituted into the equation (*) to test whether the value of Y meets the requirement that X has a real number solution
(2) When the coefficient of quadratic term is not 0, ∵ x ∈ R, ∵ Δ ≥ 0
In this case, whether the discriminant method can produce additional roots depends on whether the denominator of the equation is the same solution or not
The original problem "find the range of F (x)". The further equivalent transformation is "given that the equation y (DX ^ 2 + ex + F) = ax ^ 2 + BX + C about X has at least one real solution such that DX ^ 2 + ex + F ≠ 0, find the value range of Y."
[example]
1. When the domain of a function is a set of real numbers R
Example 1 find the range of function y = (x ^ 2-2x + 1) / (x ^ 2 + X + 1)
Because x ^ 2 + X + 1 = (x + 12) ^ 2 + 34 > 0, the domain of the function is r
Remove the denominator: Y (x ^ 2 + X + 1) = x ^ 2-2x + 1, and move to (Y-1) x ^ 2 + (y + 2) x + (Y-1) = 0
(1) When y ≠ 1, 0 ≤ y ≤ 4 is obtained from △ 0;
(2) When y = 1, we substitute it into the equation (*) to get x = 0
In conclusion, the range of the original function is [0,4]
2. When the domain of a function is not a real number set R
Example 2 find the range of function y = (x ^ 2-2x + 1) / (x ^ 2 + X-2)
It is known from the fact that the denominator is not zero that the domain of function a = {x | x ≠ - 2 and X ≠ 1}
Remove the denominator: Y (x ^ 2 + X-2) = x ^ 2-2x + 1, and sort out (Y-1) x ^ 2 + (y + 2) x - (2Y + 1) = 0
(1) When y ≠ 1, y ^ 2 ≥ 0 & is obtained from △ 0; y ∈ R
Test: y = 0 is obtained from △ = 0, and x = 1 is obtained by substituting y = 0 into the original equation, which is inconsistent with the original function definition field a,
So y ≠ 0
(2) When y = 1, we substitute it into the equation (*) to get x = 1, which is in contradiction with the original function domain a,
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So y ≠ 1
In conclusion, the range of the original function is {y | y ≠ 0 and Y ≠ 1}
If the solution set of quadratic inequality ax ^ 2 + BX + 2 > 0 is (- 1 / 2,1 / 3), then the value of a + B is?
The solution set of ax ^ 2 + BX + 2 > 0 is (- 1 / 2,1 / 3)
So the solution of ax ^ 2 + BX + 2 = 0 is x = - 1 / 2 or 1 / 3
According to Weida's theorem, X1 + x2 = - B / a = - 1 / 6, X1 * x2 = 2 / a = - 1 / 6
We get a = - 12, B = - 2
So a + B = - 14