## Related questions with answers

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

Solution

Verified$2x^3-9x^9+12x-k-0\rightarrow (1)$

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

i.e.

$2(x^2 - 2ax + a^2)(x-b) = 2x^3 - 4ax^2 + 2a^2x - 2x^2 + 4abx - 2a^2b$

$= 2x^3 - 2(2a+b)x^2 + 2(a^2 + 2ab)x + 2(a^2b)\rightarrow (2)$

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

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

Solving (3) for $b$

$b=9/2-2a$

substituting into $(4)$

$2a(a+2(9/2-2a)) = 12$

$2a(a+9-4a)=12$

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

$a(a-3) = -2$

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

$(a-2)(a-1) = 0$

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

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

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

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

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

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

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

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

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

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