请教数学paper 1

Given y=ax²+bx+c,a.b &c are postive constants. Show that the least value of y is -∆/4a where ∆=b²-4ac, & find the corresponding value of x. Sketch the graph for (1)∆>0,(2)∆<0

请问这题怎样做?

酱瓜

你不觉得第二个怪怪的吗?
它的min point是在(- , + )的哦~

不会啊...
b^2-4ac < 0 no real root 啊...
不能cut x-axis..

可是..(- , - )
两个都是-ve value哦~
真的可以酱做的啊?

lonely_world 写道:可是..(- , - )
两个都是-ve value哦~
真的可以酱做的啊?

b)∆<0

最后的min. pt is ( - b/(2a) , - ∆/(4a) ),

可是你别忘了刚才的题目是∆<0 ，也就是说 ∆ is a NEGATIVE VALUE.


所以，min. pt is ( - b/(2a) , - ∆/(4a) )
= ( - b/(2a) , - NEGATIVE VALUE/(4a) ) [Since a,b,c > 0]
= ( -ve value, -(-)/(4a) )
= ( -ve value, +ve value )


他的graph (b)是对的， min pt. is ( - b/(2a) , - ∆/(4a) )= ( -ve value, +ve value )

又有疑问了..
the roots of the equation x^2+px+q=0 differ by 4√5 and the sum of the squares of the roots is 58.Find p & q.
有人会吗?

SOR : a+b = -p --------- (1)
POR : ab =q --------- (2)

Given that
|a-b| =4√5 --------- (3)
a² + b² = 58 --------- (4)

From (3),
(|a-b|)² = (4√5)²
a²+b² -2ab = 80 -------- (5)

Substitute (2) and (4) into (5)
a²+b² -2ab = 80

58 - 2q = 80
q = -11 ------ (6)

From (1),
a+b = -p
(a+b)² = (-p)²
a² + b² + 2ab = p²
58 + 2q = p²
58 + 2(-11) = p²
p = √36
p = 6 or -6

Therefore, p= 6 and q= -11 or p=-6 and q= -11.

show that the equation f(x)=(k+1)x^2+(2k+3)x+(k+2)=0 has real roots for all real values of k. find the set of values of k for which
(a) f(x)=0 is a quadratic equation.
(b) f(x)=0 has one positive & one negative real roots.
Sketch the graph of f(x)=(k+1)x^2+(2k+3)x+(k+2) for k=-1 1/2.

你的f(x)=0是不是(k+1)x^2 + (2k+3)x+(k+2) = 0 ?
不过感觉上好像这问题是有问题。

Since f(x)=(k+1)x^2+(2k+3)x+(k+2)=0 has real roots for all real values of k,
then b^2 - 4ac > 0
=> 1>0 [我很少面对k突然间不见了，通常k还是会存在。]

has real roots(consists of different real roots & equal real roots), 应是 b^2-4ac ≥ 0
but, we get 1 > 0 就是说 different real roots ,和问题有点不符。

之前我有点混淆，起初我觉得(a)和(b)是分开来找，其实是可以同一时间利用他们来解答。

show that the equation f(x)=(k+1)x^2+(2k+3)x+(k+2)=0 has real roots
[f(x)=0 has real roots 意思是b^2 - 4ac ≥ 0 就是
equals to 0 or greater than 0 i.e.
equal real roots or distinct real roots]
for
all real values of k. find the set of values of k for which

f(x)=0 is a quadratic equation and has one positive & one negative real roots.
[意思是f(x) has two distinct real roots with one 1 +ve and 1 -ve real roots]

Sketch the graph of f(x)=(k+1)x^2+(2k+3)x+(k+2) for k=-1 1/2.


Solution：

f(x) = (k+1)x^2 + (k+3)x + (k+2) = 0

b^2 - 4ac = (k+3)^2 - 4(k+1)(k+2)
= 1 (Since b^2 - 4ac > 0,
then it has two distinct real roots)

f(x) = 0 => (k+1)x^2 + (k+3)x + (k+2) = 0

x = [-b ± √(b^2 - 4ac)]/2a
= [-(2k+3) ± √(1)]/2(k+1)
= -1 or -2(K+2)/2(k+1)
= -1 or -(K+2)/(k+1)

The two distinct real roots are -1 and -(K+2)/(k+1).
Since f(x) = 0 has one positive and one negative roots, then
x=-1 is negative root but x=-(K+2)/(k+1) is positive root.

Therefore, -(K+2)/(k+1)>0
(K+2)/(k+1)≤0
-2 ≤ k < -1

Sketch the graph of f(x)=(k+1)x^2+(2k+3)x+(k+2) for k=-1 1/2.

Solution:


Substitute k=-1 1/2 into f(x), then f(x) = -1/2 x^2 + 1/2

When f(x) = 0, then the distinct roots of f(x)= 0 are x = ±√1.
When x= 0 , then f(0) = 1/2 [It is y-intercept]


Find the turning point,
f(x) = -1/2 x^2 + 1/2
f'(x) = - x = 0 => x=0, and y = 1/2

f "(x) = -1 < 0 (It is a max. point)
Therefore, (0,1/2) is a max. point)


The graph passes through points (+√1,0), (-√1,0), and (0,1/2).
Sketch the graph based on those points with 'n' shape.

i see,real roots ---> different OR equal real roots
Q1) 为什么 -2 ≤ k ＜ 1 ？
不是 -2 < k < 1 ? ，
if k= -2 ， get 0
but, 0 ≠ POSITIVE root

Q2) for (a), k ∈R ，k≠ -1 ？

不好意思，
我忘了考虑Zero itself is neither positive nor negative.
Set of Positive Real Roots x is R+ = {x: x∈R, x>0}.

这以下是另个东西，在大学才碰到，就是说zero can be non-negative and also can be non-positive.

The non-negative numbers are the numbers that are not negative (they are positive or zero).
The non-positive numbers are the numbers that are not positive (they are negative or zero).

Solution：

f(x) = (k+1)x^2 + (k+3)x + (k+2) = 0

b^2 - 4ac = (k+3)^2 - 4(k+1)(k+2)
= 1 (Since b^2 - 4ac > 0,
then it has two distinct real roots)

f(x) = 0 => (k+1)x^2 + (k+3)x + (k+2) = 0

x = [-b ± √(b^2 - 4ac)]/2a
= [-(2k+3) ± √(1)]/2(k+1)
= -1 or -2(K+2)/2(k+1)
= -1 or -(K+2)/(k+1)

The two distinct real roots are -1 and -(K+2)/(k+1).
Since f(x) = 0 has one positive and one negative roots, then
x=-1 is negative root but x=-(K+2)/(k+1) is positive root.

Therefore, -(K+2)/(k+1)≥0
(K+2)/(k+1)<0
-2 < k < -1

Sketch the graph of f(x)=(k+1)x^2+(2k+3)x+(k+2) for k=-1 1/2.


Solution:

Substitute k=-1 1/2 into f(x), then f(x) = -1/2 x^2 + 1/2

When f(x) = 0, then the distinct roots of f(x)= 0 are x = ±√1.
When x= 0 , then f(0) = 1/2 [It is y-intercept]


Find the turning point,
f(x) = -1/2 x^2 + 1/2
f'(x) = - x = 0 => x=0, and y = 1/2

f "(x) = -1 < 0 (It is a max. point)
Therefore, (0,1/2) is a max. point)

The graph passes through points (1,0), (-1,0), and (0,1/2).
Sketch the graph based on those points with 'n' shape.


(a) 为什么你说 k ∈R ，k≠ -1 ？

问什么你觉得k≠ -1？

你要记得 N subsets Z subsets Q subsets R。
-1∈N and also ∈R.

(a) for f(x)=0 to be a quadratic equation,
coefficient of x^2 ≠ 0，
k+1≠0
k≠ -1
∴ k∈R, k≠-1

起初我不明白你问什么，所以我就以为你该不会问没有关系这题目的问题吧？
结果我就答 k ∈R ，k = -1。

不过，我忘了解答问题(a),
还是你比较细心，我人老了...

for f(x)=0 to be a quadratic equation,
coefficient of x^2 ≠ 0，
k+1≠0
k≠ -1
∴ k∈R, k≠-1

没有啦 我也是很粗心的

你明年大考是吗？

是咯！
physics and maths t我还可以，唯独pa & chem我比较担心。。

放心还有时间赶上的~
^^
加油哦！

show that 1 is a zero of the polynomial f(x)=x^3 -7x^2 +7x -11. hence, find the polynomial g(x) such that f(x)=(x-1)g(x).
show that g(x) is always positive.hence, determine the set of values of x for which x^2 +17 ≤ 7x + (11/x)

[0<x ≤1]

麻烦教我这题..

不好意思哦 lonely_world, 你确定问题有抄对吗？ 还是

f(x)=x^3 -7x^2 +7x -1

f(x)=x^3-7x^2+17x-11
就能做到了

两个都能

不过看回下部分的问题，LOG　是对的　f(x)=x^3-7x^2+17x-11