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One of the roots of 2x^3-9x^2+13x+k=0 is x=2.Find the other 2 roots. Explain how you got the answer. 

 Feb 17, 2021

Best Answer 

 #1
avatar+11656 
+5

One of the roots of 2x^3-9x^2+13x+k=0 is \(x_1=2\).Find the other 2 roots. Explain how you got the answer. 

 

Hello whatdoiputhere!

 

\((2x ^ 3-9x ^ 2 + 13x + k) \)  \(:(x-2)=\)  \(2x^2-5x+3\)

 \(\underline{2x^3-4x^2}\)

          \(-5x^2+13x\)

          \(\underline{-5x^2+10x}\)

                           \(3x+k\)

                           \(\underline{3x-6}\)

                                     \(0\)

                      

\(k=-6\)

\(2x^2-5x+3=0\\ x^2-2.5x+1.5=0\\ x=1.25\pm \sqrt{1.25^2-1.5}\\ \color{blue}x=1.25\pm0.25\)

 

\(x\in\{1,1.5,2\}\)

laugh  !

 Feb 17, 2021
edited by asinus  Feb 17, 2021
 #1
avatar+11656 
+5
Best Answer

One of the roots of 2x^3-9x^2+13x+k=0 is \(x_1=2\).Find the other 2 roots. Explain how you got the answer. 

 

Hello whatdoiputhere!

 

\((2x ^ 3-9x ^ 2 + 13x + k) \)  \(:(x-2)=\)  \(2x^2-5x+3\)

 \(\underline{2x^3-4x^2}\)

          \(-5x^2+13x\)

          \(\underline{-5x^2+10x}\)

                           \(3x+k\)

                           \(\underline{3x-6}\)

                                     \(0\)

                      

\(k=-6\)

\(2x^2-5x+3=0\\ x^2-2.5x+1.5=0\\ x=1.25\pm \sqrt{1.25^2-1.5}\\ \color{blue}x=1.25\pm0.25\)

 

\(x\in\{1,1.5,2\}\)

laugh  !

asinus Feb 17, 2021
edited by asinus  Feb 17, 2021

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