Given the function f (x) = x2-1, what is the zero point of F (x-1)

Given the function f (x) = x2-1, what is the zero point of F (x-1)

f(x-1)=(x-1)²-1=0
x-1=±1
So the zeros are x = 0 and x = 2
The correct answer is 0.2 positive solution!!!
It's on the answer
X = ± 1: friend, is there a process? Your answer is different from others
a. B is the unit vector perpendicular to each other, and | C | = 13, C.A = 3, C.B = 4,
Find the minimum of | c-t1a-t2b |
It's conditional
|c-t1a-t2b|^2
=|c|^2-6t1-8t2+t1^2+t2^2
=144+(t1-3)^2+(t2-4)^2
≥144.
If and only if T1 = 3, T2 = 4, the equal sign is taken
The minimum value of | c-t1a-t2b | is 12
It is known that the quadratic equation of one variable x ^ 2-x-k ^ 2 = 0 (k is a constant)
1) Proof: the equation has two unequal real roots;
2) Let x 1, x 2 be the two real roots of the equation, and x 1 + 2x 2 = k, then find the value of K
　　1)∵△=b²-4ac=(-1)²-4x1x(-k²)=4k²+1＞0
The equation has two unequal real roots;
2)、∵x1+x2=-b/a=1 ;x1x2=c/a=-k² x1+2x2=k
∴x1=1-x2 x2(1-x2)=-k² 1+x2=k; x2=k-1
(k-1)(2-k)=-k ²
-3k+2=0
k=-2/3
1) Proof: discriminant △ = (- 1) ^ 2-4 * 1 * (- K ^ 2)
=1+4k^2.
Corresponding to any real number k, △ = 1 + K ^ 2 > 0 is always true, so the original equation has two unequal real number roots.
2) From Veda's theorem: X1 + x2 = 1 (1)
x1*x2=-k^2. (2)
Let X1 + 2x2 = K. (3).
Expand (3)
1) Proof: discriminant △ = (- 1) ^ 2-4 * 1 * (- K ^ 2)
=1+4k^2.
Corresponding to any real number k, △ = 1 + K ^ 2 > 0 is always true, so the original equation has two unequal real number roots.
2) From Veda's theorem: X1 + x2 = 1 (1)
x1*x2=-k^2. (2)
Let X1 + 2x2 = K. (3).
Divide (3) into X1 + x2 + x2 = K
∵x1+x2=1. ∴1+x2=k, x2=k-1. (4）。
Substituting (4) into (2), we get the following result:
x1*(k-1)=-k^2.
x1=-k^2/(k-1).
x1+x2=-k^2/(k-1)+k-1=1.
After removing the denominator and simplifying, we get: - K ^ 2 + (k-1) ^ 2 = k-1
-k^2+k^2-2k+1=k-1.
3k=2,
That is to say, k = 2 / 3. Put it away
How to solve x + 3x = - 16 by shifting and merging similar terms
X + 3x + 16 = 0
If we combine the similar items, we get 4x + 16 = 0
If the coefficient is changed to 1, x + 4 = 0
X=-4
Factorization: A ^ 3b-4a ^ 2B ^ 2 + 4AB ^ 3
Given the function f (x) = 1 + x-x2 / 2 + X3 / 3 - +X ^ 2013 / 2013, Let f (x) = f (x + 4), and the zeros of function f (x) are in [a, b] (a < B, a, B ∈ z), then the minimum value of B-A is?
Given a (- 2,4) B (3, - 1) C (- 3, - 4), and vector cm = 3, vector Ca, vector CN = 2, vector CB, find the coordinates of M, N and vector Mn
Let m (x, y) cm = (x + 3, y + 4) cm = 3CA = 3 (1,8) = (3,24) x + 3 = 3, y + 4 = 24, the solution is x = 0, y = 20, m (0,20) let n (x1, Y1) CN = (x1 + 3, Y1 + 4) CN = 2CB = 2 (6,3) = (12,6) X1 + 3 = 12, Y1 + 4 = 6, the solution is X1 = 9, Y1 = 2, n (9,12) vector Mn = (9, - 8)
Let m (x, y) cm = (x + 3, y + 4)
CM=3CA=3(1,8)=(3,24)
X + 3 = 3, y + 4 = 24, x = 0, y = 20 m (0,20)
Let n (x1, Y1) CN = (x1 + 3, Y1 + 4)
CN=2CB=2(6,3)=(12,6)
X 1 + 3 = 12, Y 1 + 4 = 6, x 1 = 9, Y 1 = 2, n (9,2)
Vector Mn = (9, - 18)
Given that one root of the quadratic equation x & # 178; - 6x + k = 0 is the square of the other root, find the value of K
Let's say the two heels are X1 and x2
So X1 + x2 = 6
Suppose X1 = x2 ^ 2
So x ^ 2 + x2 = 6
x^2+x2-6=0
(x2-2)(x2+3)=0
X2 = 2, X1 = 4, then k = X1 * x2 = 8
perhaps
X2 = - 3, X1 = 9, then k = X1 * x2 = - 27
Let the root be a, a ^ 2, from the relationship between root and coefficient, a + A ^ 2 = 6, get a, and then from a * a ^ 2 = a ^ 3 = K
According to Weida's theorem, X1 + x2 = 6, X1 * x2 = K
So X1 + (x1) ^ 2 = 6, that is, (x1) ^ 2 + x1-6 = (x1-2) (x1 + 3) = 0
So X1 = 2 or X1 = - 3
The corresponding x2 = 4 or x2 = 9
So k = 8 or K = - 27
Let these two roots be a and a ^ 2
A + A ^ 2 = 6 can be obtained from Veda's theorem
The solution is a = - 3 or a = 2
So two of the equations are 2 and 4 or - 3 and 9
From Veda's theorem, we can get k = 2 × 4 = 8 or K = - 3 × 9 = - 27
How about 16 × 5 + 3 / 5 x = x,
16 × 5 + 3 / 5 x = x
80=X-3/5X
2/5X=80
X=80÷2/5
X=200
If (a-b) (a + 2b) = A & # 178; + mAb + Nb & # 178;, then M + n=
（a-b）（a+2b）=a²+mab+nb²
Namely
a²＋ab-2b²=a²+mab+nb²
therefore
m=1,n=-2
m＋n=1-2=-1
The original formula = (a-b) [(a + b) + b] = A & # 178; - B & # 178; + ab-b & # 178; = A & # 178; + ab-2b & # 178;
That is, M = 1, n = - 2
m+n=-1
(a-b)(a+2b)=a²+mab+nb²
a²+ab-2b²=a²+mab+nb²
M=1
n=-2
m+n=1-2=-1