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If two of the roots of 2x^3 + 8x^2 - 120x + k = 0 are equal, find the value of k given that k is positive.

 Aug 22, 2019
 #1
avatar+106535 
+1

2x^3 + 8x^2  - 120x + k  =  0     divide through by  2

 

x^3 + 4x^2  - 60x  + k/2  = 0    (1)

 

Let the two equal roots   = m      and let the other root be n

 

So we have

 

(x - m)^2 (x - n)   = 0

 

(x^2 - 2xm + m^2) ( x - n)  =  0

 

x^3 -2mx^2 + m^2x - x^2n + 2mnx -m^2n  = 0

 

x^3 - (2m + n)x^2 + (m^2+2mn)x -m^2n  = 0      (2)

 

Equating coefficients  in (1)  and (2)   we have the following system

 

-(2m + n)  =  4    ⇒   2m + n  = -4    ⇒  n =  -4 - 2m      ( 3)

(m^2 +2mn) = -60       (4)

-m^2n = k/2    ⇒  -2m^2n  = k   (5)

 

Sub  (3)  into (4)  for  n    and we have that

 

m^2  + 2m (-4 -2m) = -60

 

m^2  - 8m - 4m^2  = -60

 

-3m^2 -8m + 60  = 0     

 

3m^2 + 8m  - 60  = 0      factor

 

(3m  - 10) ( m + 6)  = 0

 

Set each factor to 0  and solve for m and we have that   m  =10/3   or  m = -6

 

If m  = -6   then n =  -4 - 2(-6)  = 8     and   using (5) gives  -2(-6)^2 * 8  = -576  = k     [reject]

 

If m = 10/3   then n  = -4 - 2(10/3)  = -32/3  and  (5) gives  -2 (10/3)^2* (-32/3) =  6400/27 = k

 

 

cool cool cool

 Aug 22, 2019
 #2
avatar+23884 
+2

If two of the roots of \(2x^3 + 8x^2 - 120x + k = 0\) are equal,

find the value of k given that k is positive.

 

An alternative solution:

so \(y=0\) and \(y'=0\)

 

\(\begin{array}{|rcll|} \hline y = 0 &=& 2x^3 + 8x^2 - 120x + k \\ y'= 0 &=& 6x^2 + 16x - 120 \\ \hline 6x^2 + 16x - 120 &=& 0 \quad & | \quad : 2 \\ 3x^2 + 8x - 60 &=& 0 \\\\ x&=& \dfrac{ -8\pm \sqrt{64-4\cdot 3\cdot(-60)} } {2\cdot 3} \\ x&=& \dfrac{ -8\pm \sqrt{64+720} } {6} \\ x&=& \dfrac{ -8\pm \sqrt{784} } {6} \\ x&=& \dfrac{ -8\pm 28 } {6} \\\\ x_1 &=& \dfrac{ -8+ 28 } {6} \\ x_1 &=& \dfrac{20} {6} \\ \mathbf{x_1} &=& \mathbf{\dfrac{10} {3}} \\\\ x_2 &=& \dfrac{ -8- 28 } {6} \\ x_2 &=& \dfrac{-36} {6} \\ \mathbf{x_2} &=& \mathbf{-6} \\ \hline \end{array} \)

 

\(\begin{array}{|l|rcll|} \hline x_1 = \dfrac{10} {3}: & 2\cdot \left(\dfrac{10} {3}\right)^3 + 8\cdot \left(\dfrac{10} {3}\right)^2 - 120\cdot \left(\dfrac{10} {3}\right) + k &=& 0 \\ & -237.\overline{037} + k &=& 0 \\ & \mathbf{k} &=& \mathbf{237.\overline{037}} \ \checkmark\quad & | \quad k>0! \\ \hline \end{array} \)

 

\(\begin{array}{|l|rcll|} \hline x_2 =-6: & 2 \left(-6\right)^3 + 8 \left(-6\right)^2 - 120 \left(-6\right) + k &=& 0 \\ & -432+288+720+ k &=& 0 \\ & 576+ k &=& 0 \\ & \mathbf{k} &=& \mathbf{-576} \quad & | \quad k<0\ \text{ no solution!}\\ \hline \end{array} \)

 

laugh

 Aug 23, 2019

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