F4 Math一問~~~~~

a. (m - 1)x2 + 2x + (2m - 3) = 0

Since the equation has two equal roots,

So, discriminant = 0

(2)2 – 4(m - 1)(2m - 3) = 0

4 – 4(2m2 – 5m + 3) = 0

1 – 2m2 + 5m – 3 = 0

2m2 – 5m + 2 = 0

(2m - 1)(m - 2) = 0

m = 2 or 1/2

b. For m = 2, the equation becomes

x2 + 2x + 1 = 0

(x + 1)2 = 0

x = -1

For m = 1/2, the equation becomes,

-x2/2 + 2x – 2 = 0

x2 – 4x + 4 = 0

(x - 2)2 = 0

x = 2

(m - 1)x^2 + 2x + (2m - 3) = 0
a ) Find the possible values of m .
When the quadratic equation has two equal roots ,
Δ=0,
b^2 - 4ac = 0
(2)^2 - 4(m - 1)(2m - 3) = 0
4 - 4(m - 1)(2m - 3) = 0
4 - 4(2m^2 - 5m + 3) = 0
4 - 8m^2 + 20m - 12 = 0
8m^2 - 20m + 8 = 0
(4m - 2)(2m - 4) = 0
m = 1/2 or m = 2
b ) For each value of m , solve the equation .
when m = 2 ,
(2 - 1)x^2 + 2x + [2(2) - 3] = 0
x^2 + 2x + 1 = 0
(x+1)^2 = 0
x = -1
when m = 1/2 ,
(1/2 - 1)x^2 + 2x + [2(1/2) - 3] = 0
-1/2x^2 + 2x - 2 = 0
-x^2 + 4x - 4 =0
x^2 - 4x +4 = 0
(x - 2)^2 = 0
x = 2