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.
+128474 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
+26367 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 \(y'=0\)
\(\begin{array}{|rcll|} \hline y = 0 &=& 2x^3 + 8x^2 - 120x + k \\ y'= 0 &=& 6x^2 + 16x - 120 \\ \hline 6x^2 + 16x - 120 &=& 0 \quad & | \quad : 2 \\ 3x^2 + 8x - 60 &=& 0 \\\\ x&=& \dfrac{ -8\pm \sqrt{64-4\cdot 3\cdot(-60)} } {2\cdot 3} \\ x&=& \dfrac{ -8\pm \sqrt{64+720} } {6} \\ x&=& \dfrac{ -8\pm \sqrt{784} } {6} \\ x&=& \dfrac{ -8\pm 28 } {6} \\\\ x_1 &=& \dfrac{ -8+ 28 } {6} \\ x_1 &=& \dfrac{20} {6} \\ \mathbf{x_1} &=& \mathbf{\dfrac{10} {3}} \\\\ x_2 &=& \dfrac{ -8- 28 } {6} \\ x_2 &=& \dfrac{-36} {6} \\ \mathbf{x_2} &=& \mathbf{-6} \\ \hline \end{array} \)
\(\begin{array}{|l|rcll|} \hline x_1 = \dfrac{10} {3}: & 2\cdot \left(\dfrac{10} {3}\right)^3 + 8\cdot \left(\dfrac{10} {3}\right)^2 - 120\cdot \left(\dfrac{10} {3}\right) + k &=& 0 \\ & -237.\overline{037} + k &=& 0 \\ & \mathbf{k} &=& \mathbf{237.\overline{037}} \ \checkmark\quad & | \quad k>0! \\ \hline \end{array} \)
\(\begin{array}{|l|rcll|} \hline x_2 =-6: & 2 \left(-6\right)^3 + 8 \left(-6\right)^2 - 120 \left(-6\right) + k &=& 0 \\ & -432+288+720+ k &=& 0 \\ & 576+ k &=& 0 \\ & \mathbf{k} &=& \mathbf{-576} \quad & | \quad k<0\ \text{ no solution!}\\ \hline \end{array} \)
