諗唔到點計呀,,

if the equation x^2-2(k-5) x +16 =0 have a double real roots
a ....find the values of k
x^2 - 2(k-5) x + 16 = 0
判別式
= [-2(k-5)]^2 - 4(16)
= 4(k^2 - 10k + 25) - 64
= 4(k^2 - 10k + 9)
double real roots, 4(k^2 - 10k + 9) = 0
k^2 - 10k + 9 = 0
(k - 1)(k - 9) = 0
k = 1 or k = 9
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b...the root of the equation for each value of k
when k = 1,
x^2 - 2(k-5) x + 16 = 0
x^2 + 8 x + 16 = 0
(x + 4)^2 = 0
x = -4
when k = 9,
x^2 - 8 x + 16 = 0
(x - 4)^2 = 0
x = 4

a) ∆=0=b^2-4ac
0=〖[-2(k-5)]〗^2-4(1)(16)
0=〖4k〗^2-40k+36
0=k^2-10k+9
0=(k-1)(k-9)
k=1 or k=9

b) x^2-2(k-5)x+16=0……(1)
Put k = 1 into (1)
x^2-2(1-5)x+16=0
x^2+8x+16=0
(x+4)^2=0
x=-4

Put k = 9 into (1)
x^2-2(9-5)x+16=0
x^2-8x+16=0
(x-4)^2=0
x=4