math急 20點!!!!!!!

Question

1.consider the quadratic equation (k+2)x^2+(3-k)x-4=0,where k is not equal -2.find the value of k if

(a)one root is 2

(b)the product of the roots is 6

(c)the roots are equal in magnitude but opposite in sign

(d)one root is the reciprocal of the other root

2.α and β are the roots of x^2-7x+k=0.If (α-β)^2+9 find the value of k

function

3.let f(x) +x^2+4x-2 and g(x)=2f(x+1)-f(x).

(a)find g(x)

(b)g(w)=-8 fond the value of w

Answer 1

最佳解答

預備知識 : 根與係數關係

如果α and β are the roots of ax^2+bx+c=0

那麼ax^2+bx+c=a(x-α)(x-β)=ax^2-a(α+β)x+aαβ=0

所以α+β=-b/a αβ=c/a

1.consider the quadratic equation (k+2)x^2+(3-k)x-4=0,where k is not equal -2.find the value of k if

(a)one root is 2

(b)the product of the roots is 6

(c)the roots are equal in magnitude but opposite in sign

(d)one root is the reciprocal of the other root

sol:

(a)x以2代入

4k+8+6-2k-4=0 k=-5#

(b)the product of the roots is 6

-4/(k+2)=6 6k+12=-4 k=-8/3#

(c)the roots are equal in magnitude but opposite in sign

所以兩根和-(3-k)/(k+2)=0 k=3#

(d)one root is the reciprocal of the other root

猜測是兩根互為倒數 其積=1

-4/(k+2)=1 k=-6#

2.α and β are the roots of x^2-7x+k=0.If (α-β)^2+9 find the value of k

sol:

我猜測題目打字錯誤 修改為......If (α-β)^2=9 ,find ......

α+β=7 而且(α+β)^2-(α-β)^2 =4αβ

49-9=4αβ αβ=10

k/1=10 k=10#

3.let f(x)修改為=x^2+4x-2 and g(x)=2f(x+1)-f(x).

(a)find g(x)

(b)g(w)=-8 fond the value of w

sol:

(a) g(x)=2f(x+1)-f(x)=2[(x+1)^2+4(x+1)-2]- (x^2+4x-2)=x^2+8x+8#

(b)w^2+8w+8=-8 w=-4#

Answer 2

阿霹 + 1! Very nice, especially this part:

ax^2+bx+c=a(x-α)(x-β)=ax^2-a(α+β)x+aαβ=0