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Question

Determine the product of all values of k for which the polynomial equation 2x39x2+12xk=02 x^{3}-9 x^{2}+12 x-k=0 has a double root.

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Answered 2 years ago
Answered 2 years ago
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2x39x9+12xk0(1)2x^3-9x^9+12x-k-0\rightarrow (1)

To have a double root the equation would be 2(xa)2(xb)2(x-a)^2(x-b) where aa is the double root.

i.e.

2(x22ax+a2)(xb)=2x34ax2+2a2x2x2+4abx2a2b2(x^2 - 2ax + a^2)(x-b) = 2x^3 - 4ax^2 + 2a^2x - 2x^2 + 4abx - 2a^2b

=2x32(2a+b)x2+2(a2+2ab)x+2(a2b)(2)= 2x^3 - 2(2a+b)x^2 + 2(a^2 + 2ab)x + 2(a^2b)\rightarrow (2)

Compare eqs.(1)eqs. (1) and (2)(2)

2(2a+b)=9(3),2a(a+2b)=12(4),k=2a2(b)2(2a+b)=9\rightarrow (3), 2a(a+2b)=12\rightarrow (4), k = 2a^2(b)

Solving (3) for bb

b=9/22ab=9/2-2a

substituting into (4)(4)

2a(a+2(9/22a))=122a(a+2(9/2-2a)) = 12

2a(a+94a)=122a(a+9-4a)=12

a(3a+9)=6a(-3a+9) = 6

a(a3)=2a(a-3) = -2

a23a+2=0a^2 - 3a + 2 = 0

(a2)(a1)=0(a-2)(a-1) = 0

so a=2a = 2 or a=1a = 1

b=9/22(2)=1/2b = 9/2 - 2(2) = 1/2 or b=9/22(1)=5/2b = 9/2 - 2(1) = 5/2

For a=2,b=1/2a=2, b=1/2

Eq. (3) 2(2(2)+(1/2))=2(4+1/2)=2(9/2)=92(2(2)+(1/2)) = 2(4+1/2) = 2(9/2) = 9

Eq. (4) 2(2)(2+2(1/2))=4(3)=122(2)(2+2(1/2))=4(3) = 12

For a=1,b=5/2a=1, b=5/2

Eq. (3) 2(2(1)+(5/2))=2(2+5/2)=2(9/2)=92(2(1)+(5/2))=2(2+5/2)=2(9/2) = 9

Eq. (4) 2(1)(1+2(5/2))=2(6)=122(1)(1+2(5/2))=-2(6)=12

So k=2a2(b)=2(2)2(1/2)=4k = 2a^2(b) = 2(2)^2(1/2) = 4 or k=2(12)(5/2)=5k = 2(1^2)(5/2) = 5

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