中4. maths~

1a)
　x^2 + x - k^2 - k
= (x^2 - K^2) + (x - k)
= (x + k)(x - k) + (x + k)
= (x + k)(x - k + 1)

b)
(x+4)(x-3) = (k+4)(k-3)
x^2 + x - 12 = k^2 + k - 12
x^2 + x - k^2 - k = 0
(x + k)(x - k + 1) = 0 (from a)
so, x = -k or k - 1

2a)
(x-1)^2 - k(x-2) = 0
x^2 - 2x + 1 - kx + 2l = 0
x^2 - (2 + k)x + (1 + 2k) = 0

b)
for equal roots, Δ=0,
(2 + k)^2 - 4(1)(1 + 2k) = 0
4 + 4k + k^2 - 4 - 8k = 0
k^2 - 4k = 0
k(k - 4) = 0
so, k = 0 or 4

c)
when k = 4
x^2 - 6x + 9 = 0 (from a)
(x - 3)^2 = 0
x = 3(repeated)

2007-02-23 14:58:03 補充：
sorry, 1a 同 b個符號錯左,,,應該係咁先arm!1a)　x^2 x - k^2 - k= (x^2 - K^2) (x - k)= (x k)(x - k) (x - k)= (x - k)(x k 1)b)(x 4)(x-3) = (k 4)(k-3) x^2 x - 12 = k^2 k - 12x^2 x - k^2 - k = 0(x - k)(x k 1) = 0 (from a)so, x = k or -(k 1)希望可以幫到你!

1a, Using Delta = b^2 - 4ac = 1 - 4(1) (-k^2-k) = 4k(k-1)
Hence, x = -b + Delta / 2a or -b - Delta / 2a = -1/2 + 2k(k-1) or -1/2 - 2k(k-1)
b, factorize the eqt. we have, x^2 + x - k^2 - k, From a, x = -1/2 + 2k(k-1) or -1/2 - 2k(k-1)

2a, (x-1)^2 - k(x-2) =0 => x^2 - 2x + 1 - kx -2k = x^2 - (k+2)x - (2k-1)
b, X have equal rt. i.e. Delta = 0
(k + 2)^2 + 4 (2k-1) = 0
k^2 + 12k = 0
k = 0 or -12.
c, K is greatest, i.e. k = 0
Hence, Eqt. remind as (x-1)^2= 0
x = 1 (repeated.)

1a)
Factorize x^2 + X - k^2 - k

x^2 + X - k^2 - k

= (x + k)(x - k) + (x - k)

= (x - k)(x + k + 1)


b) hence solve the equation (x+4)(x-3) = (k+4)(k-3) where k is a real constant.

(x+4)(x-3) = (k+4)(k-3)

x = k

2 consider the equation (x-1)^2 - k(x-2) = 0

a) rewrite the equation in the form x^2+bx+c =0

(x-1)^2 - k(x-2) = 0

x^2 - 2x + 1 - kx + 2k = 0

x^2 - (2 + k)x + 1 + 2k = 0

b) if the equation has equal roots, find the possible values of k.

(2 + k)^2 - 4(1 + 2k) = 0

4 + 4k + k^2 - 4 - 8k = 0

k^2 - 4k = 0

k(k - 4) = 0

k = 0 or k = 4

c)
if k takes the greater value found in (b) ,solve the given equation.

so,k = 4

x^2 - 6x + 5 = 0

(x - 5)(x - 1) = 0

x = 5 or x = 1

2007-02-23 14:50:52 補充：
1b) hence solve the equation (x 4)(x-3) = (k 4)(k-3) where k is a real constant.(x 4)(x-3) = (k 4)(k-3)x^2 x - 12 = k^2 k - 12x^2 x - k^2 - k = 0(x - k)(x k 1) = 0x = k or x = 1 - k

1a) x^2 + x - k^2 - k
= x^2 - k^2 + (x - k)
= (x - k)(x + k) + (x - k)
= (x - k)(x + k + 1)

b) By setting k = 3, lhs = (x+4)(x-3) = x^2 + x - 3^2 - 3 = x^2 + x - 12
Similarly, rhs = (k+4)(k-3) = k^2 + k - 12
Since lhs = rhs, x^2 + x - 12 = k^2 + k - 12
=> x^2 + x - k^2 - k = 0
=> (x - k)(x + k + 1) = 0
=> x = k or x = -(k+1) where k is a real constant

2a) (x-1)^2 - k(x-2) = 0
=> (x^2 - 2x + 1) - (kx -2k) = 0
=> x^2 - (2+k)x + (1+2k) = 0

b) If the equation in 2a) has equal roots, delta = (2+k)^2 - 4x1x(1+2k) = 0
=> 4 + 4k + k^2 - (4 + 8k) = 0
=> k^2 - 4k = 0
=> k(k - 4) = 0
=> k = 0 or k = 4

c) k takes the greater value found in 2b) means k = 4 is chosen.
The only root is [(2+4)+0]/2 or [(2+4)-0]/2 = 3.
{by quadratic formula: [-b(+or-)sqrt(b^2-4ac)]/2}