F4-解一元二次方程

1.x^2-7x+12=0
x=3 or x=4

2.12x^2-13x+3=0
x=1/3 or x=3/4

3.x^2-4x+2=0
x=-(-4)±√[(-4)^2-4(1)(2)]/2(1)
x=(4±√8)/2
x=2±√2
x=2+√2 or x=2-√2

4.8x^2-x-4=0
x=-(-1)±√[(-1)^2-4(8)(-4)]/2(8)
=(1±√129)/16
x=1+√129/16 or x=1-√129/16

5.4x^2+4x+k=0
Δ=0
(4)^2-4(4)k=0
16-16k=0
-16k=-16
k=1

6.14x^2-7x+k=0
Δ＞0
(-7)^2-4(14)k＞0
49-56k＞0
-56k＞-49
k＜7/8

7.y=px^2-3x-5
Δ＜0
(-3)-4(p)(-5)＜0
9+20p＜0
20p＜-9
p＜-9/20

8.x^2+4x-k=0
Δ＞0
(4)^2-4(1)(-k)＞0
16+4k＞0
4k＞-16
k＞-4

9a.(m-2)x^2+mx+2=0
Δ=0
(m)^2-4(m-2)(2)=0
m^2-8m+16=0
m=4

b.(4-2)x^2+4x+2=0
2x^2+4x+2=0
x=-1 or x=-1

Questions and Answers :

1)x^2-7x+12=0
(x-3)(x-4) = 0
x = 3 or 4 #

2)12x^2-13x+3=0
(4x-3)(3x-1) = 0
x = 3/4 or 1/3 #

3)x^2-4x+2=0
(x-2)^2 = 0
x = 2 (repeated)

4)8x^2-x-4=0
x^2 - x/8 - 1/2 =0
x^2 - x/8 = 1/2
x^2 - x/8 + (1/16)^2 = 1/2 + (1/16)^2
(x- 1/16) ^2 = 129/256
x - 1/16 = (±√129 / 16)
x = 1/16 ±√129 / 16
x = (1±√129) / 16 #

5)If the equation 4x^2+4x+k=0 has two equal real roots, find the value of k.
∵4x^2+4x+k=0 has two equal real roots
∴△=0
4^2 - 4 (4)(k) = 0
16-16k = 0
k = 1 #

1)
x² - 7x + 12 = 0
(x - 3)(x - 4) = 0
x - 3 or x = 4


=====
2)
12x² - 13x + 3 = 0
(4x - 3)(3x - 1) = 0
x = 3/4 or x = 1/3


=====
3)
x² - 4x + 2 = 0
x = [4 ± √(4² - 4*1*2)] / 2
x = 2 + √2 or x = 2 - 2√2


=====
4)
8x² - x - 4 = 0
x = [1 ± √(1² + 4*8*4)] / (2*8)
x = (1 + √129)/16 or x = (1 - √129)/16


5)
The equation 4x² + 4x + k = 0 has two equal real roots.
Then, the discriminant Δ = 0
4² - 4*4*k = 0
16 - 16k = 0
k = 1


=====
6)
The equation 14x² - 7x + k = 0 has two distinct realroots.
Then, the discriminant Δ > 0
(-7)² - 4*14*k > 0
49 - 56k > 0
56k < 49
k < 7/8


=====
7)
The graph of y = px² - 3x - 5 and the x-axis has no points of intersection.
px² - 3x - 5 = 0, discriminant Δ < 0
(-3)² - 4*p*(-5) < 0
9 + 20p < 0
20p < -9
p < -9/20


=====
8)
The equation x² + 4x - k = 0 has two distinct real roots.
Then, the discriminant Δ > 0
4² - 4*1*(-k) > 0
16 + 4k > 0
4k > -16
k > -4


=====
9)
a)
The quadratic equation (m - 2)x² + mx + 2 = 0 has two equal roots.
Then, the discriminant Δ = 0
m² - 4*(m - 2)*(2) = 0
m² - 8m + 16 = 0
(m - 4)² = 0
m = 4 (double roots)

b)
The equation is :
(4 - 2)x² + 4x + 2 = 0
x² + 2x + 1 = 0
(x + 1) = 0
x = -1 (double roots)

2012-10-04 22:49:42 補充：
9b. There is a typo.
The last second line should be : (x + 1)² = 0