a math

2006-10-26 7:04 am
1, 若8x^2-ax+9=0的其中一個根是另一個根的兩倍,求方程的根和a的值。

2, 若二次方程kx^2+4x+k^2-21=0的兩根的積是 4,求k的值。

3a, 若方程x^2+a(3a-5)x=2(x+4a)的其中一個根是另一個根的負值,求a的值。
b, 若a>0,運用(a)小題的結果解方程。

4, 若a、b是x^2-5x+k=0的根,而 1/2a+1、1/2b+1是35x^2+nx+1=0的根,求k和n的值。

5, 方程x^2+px+q=0的兩根的差為2,證明p^2=4+4q。

6, 若二次方程ax^2+bx+c=0的兩根的平方之和是4,證明b^2=2ac+4a^2。

列明步驟

回答 (1)

2006-10-26 7:55 am
✔ 最佳答案
1.Let the two roots of 8x^2-ax+9=0 be h,k

set h=2k

sum of root:h+k=a/8
product of root:hk=9/8

3k=a/8
2k^2=9/8

a=18

8x^2-18x+9=0
x=3/2 or 3/4



2.Let the two roots of kx^2+4x+k^2-21=0 be a, b

product of root:ab=4=k^2-21/k

k=7 or -3



3(a).Let the roots of x^2+a(3a-5)x=2(x+4a) be h, k

k=-h

sum of root:h+(-h)=0=-a(3a-5)

a=5/3



3(b).a=0?

then x^2+(0)(3(0)-5)x=2(x+4(0))

x^2-2x=0

x=2 or 0


4.a, b are the root of x^2-5x+k=0...(1)
1/2a+1, 1/2b+1 are the root of 35x^2+nx+1=0...(2)

From (1),

sum of root:a+b=5
product of root:ab=k

From (2),

sum of root: (1/2a+1)+(1/2b+1)=-n/35...(3)
product of root: (1/2a+1)(1/2b+1)=1/35...(4)

From (3),

(a+b)/2+2=-n/35
5/2+2=-n/35

n=-157.5

From (4),

(1/2a+1)(1/2b+1)=1/35
ab/2+(a+b)/2+1=1/35
ab/2+(5)/2+1=1/35
ab=k=-243/35


5.Let the root of x^2+px+q=0 be a, b

a-b=2

sum of root: a+b=-p
product of root: ab=q

(a+b)^2-4ab=(a-b)^2
p^2-4q=4
p^2=4+4q


6.let the two root of ax^2+bx+c=0 be h, k

h^2+k^2=4

sum of root: h+k=-b/a
product of root: hk=c/a

(h+k)^2-2hk=h^2+k^2=4
(-b/a)^2-2(c/a)=4
b^2-2ac=4a^2
b^2=2ac+4a^2

下次咁多題目
較高分D啦


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