幾條中四的數

1. f(x)=x^2+2kx-1

f(-1)=f(1/2)
(-1)^2 + 2k(-1) -1 = 1/2^2 + 2k(1/2) -1
1 - 2k = 1/4 + k
3k = 3/4
k = 1/4

2. 6x^2+4x-n+4=0 has no real roots

discriminate < 0

4^2 - 4(6)(-n+4) < 0
16 + 24n - 96 < 0
n < 10/3

3. (a+b)x^2-2ax+(a-b)=0
disciminate = (-2a)^2 - 4(a+b)(a-b)
= 4(a^2) - 4(a^2) + 4(b^2)
= 4(b^2) where b is a real no.
discriminate > 0
So (a+b)x^2-2ax+(a-b)=0 has real roots

4. (x^2-5)(x+4)-(2x^2-9)≡x^3+Ax^2-(2A+B)x-11
LHS = (x^2-5)(x+4)-(2x^2-9)
= x^3 + 4x^2 - 5x -20 - 2x^2 + 9
= x^3 + 2x^2 -5x - 11

By comparing coeff.,
A = 2

-(2A + B) = -5
2(2) + B = 5
B = 1

To do these questions, you should understand the followings:

For parabola f(x) = Ax^2 Bx C
The discriminant is B^2 - 4AC, and it is used to determine the number of roots the parabola f(x) has.

If f(x) has 2 real roots, then the B^2 - 4AC > 0,
If f(x) has 1 real root, then the B^2 - 4AC = 0,
If f(x) has no real roots, then the B^2 - 4AC < 0

Now, to your questions,
Q1. 設f(x)=x^2 2kx-1,若f(-1)=f(1/2),求k

Sub x = -1 into f(x)

f(-1) = (-1)^2 2k(-1) - 1
= 1 - 2k - 1 = -2k

Sub x = 1/2 into f(x)

f(1/2) = (1/2)^2 2k(1/2) - 1
= 1/4 k - 1

= k - 3/4

Now, since f(-1)=f(1/2)

Therefore, -2k = k - 3/4
3k = 3/4
k = 1/4

Q2. 已知二次方程6x^2 4x-n 4=0沒實根,求n

Let f(x) = 6x^2 4x-n 4,

If f(x) does not have any real root, then the discriminant b^2 - 4ac < 0

In f(x),
a = 6
b = 4
c = 4-n

4^2 - 4 x 6 x (4-n) < 0
16 - 24(4-n) < 0
16 - 96 24n < 0
24n < 80
n < 10/3

Q3. 設a和b為兩實數,證明二次方程(a b)x^2-2ax (a-b)=0必有實根

If f(x) = (a b)x^2-2ax (a-b)=0,

To determine the parabola has root, let's examine the discriminant B^2 - 4AC

A = (a b)
B = -2a
C = (a-b)

The discriminant = B^2 - 4AC = (-2a)^2 - 4(a b)(a-b)

= 4a^2 - 4(a^2-b^2)
= 4a^2 - 4a^2 4b^2
= 4b^2 or (2b)^2

Since the square of any real number is bigger or equal to 0, therefore (2b)^2 >= 0, and therefore the discriminant is bigger or equal to 0.

Therefore, the parabola f(x) has at least 1 root.

Q4. 因式分解
若有若無(x^2-5)(x 4)-(2x^2-9)≡x^3 Ax^2-(2A B)x-11

(x^2-5)(x 4)-(2x^2-9) ≡ x^3 Ax^2-(2A B)x-11

LHS = (x^3 4x^2 -5x -20) - (2x^2 - 9)
= x^3 4x^2 -5x - 20 -2x^2 9
= x^3 2x^2 -5x - 11

RHS = x^3 Ax^2-(2A B)x-11

By comparing the co-efficient of the LHS and RHS,

A = 2 and 2A+B = 5

and B = 1