一元二次方程(3)

1) the coordinate of point at x-intercept is (x, 0)
therefore ,
0 = -3x^2 +-1
3x^2+-1 = 0
3x^2 + 1 = 0 or 3x^2 - 1 = 0
x^2 = -1/3 (rejected)

therefore 3x^2 - 1 = 0
x^2 = 1/3
x = sqrt(1/3) or - sqrt(1/3)

2) 2x^2-4x+k=0
discriminant < 0
(-4)^2 - 4(2)(k) < 0
16 - 8 k< 0
2 - k < 0
2< k
k > 2

3) x^2+2x-kx+10-k=0
discriminant > 0
(2-k)^2 - 4(10-K) > 0
4 - 4k + k^2 - 40 + 4k > 0
-36 + k^2 > 0
k^2 > 36
k > 6 or k < -6

4) 2x(x-1)=k-5
2x^2 - 2x -k + 5 = 0
discriminant < 0
(-2)^2 - 4(2)(5-k) < 0
4 - 8(5-k) < 0
4 - 40 + 8k < 0
-36 + 8k < 0
8k < 36
k < 36/8
k < 9/2 = 4.5
becasue k is an integer so the nearest integer is 4

5) y=x^2+k(x-1)+9-x
= x^2 +kx - k + 9 - x
= x^2 + (k-1)x + 9 - k
From the equation above, the discriminant = 0
(k-1)^2-4(9-k) = 0
k^2-2k+1-36+4k = 0
k^2+2k-35 = 0
k = -2+-sqrt(2^2-4(-35))/2
= (-2+-12)/2
= (-2 + 12)/2 or (-2 - 12)/2
= 5 or -7

1) x = 1/Sqrt3 or x = -1/Sqrt3

2) k > 2

3) k > 6, k < -6

4) k < 9/2 but k is an integer so k = 4

5) k = 4Sqrt2 or k = -4Sqrt2