Quadratic Polynomial Help! (1 Viewer)

[JasonNg]

Hey guys! Can you guys please help me with this question?

The roots of the quadratic equation x^2 - (2k+4)x + (k^2 + 3k + 2) = 0 are non zero and one is twice the other root. Calculate the value of k.

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Let the roots be a and 2a. Using sum and product of roots...}\\ (1)- 3a=2k@plus;4 \\ (2)- 2a^2=k^2@plus;3k@plus;2 \\ \textup{Subbing the value of a from (1) }\\ 2(\frac{2k@plus;4}{3})^2=k^2@plus;3k@plus;2\\ 2(4k^2@plus;16k@plus;16)=9k^2@plus;27k@plus;18\\ k^2-5k-14=0\\ (k-7)(k@plus;2)=0 \\ \textup{k=-2 or k=7} \\\textup{But the roots are non-zero. Therefore k=7}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Let the roots be a and 2a. Using sum and product of roots...}\\ (1)- 3a=2k+4 \\ (2)- 2a^2=k^2+3k+2 \\ \textup{Subbing the value of a from (1) }\\ 2(\frac{2k+4}{3})^2=k^2+3k+2\\ 2(4k^2+16k+16)=9k^2+27k+18\\ k^2-5k-14=0\\ (k-7)(k+2)=0 \\ \textup{k=-2 or k=7} \\\textup{But the roots are non-zero. Therefore k=7}" title="\textup{Let the roots be a and 2a. Using sum and product of roots...}\\ (1)- 3a=2k+4 \\ (2)- 2a^2=k^2+3k+2 \\ \textup{Subbing the value of a from (1) }\\ 2(\frac{2k+4}{3})^2=k^2+3k+2\\ 2(4k^2+16k+16)=9k^2+27k+18\\ k^2-5k-14=0\\ (k-7)(k+2)=0 \\ \textup{k=-2 or k=7} \\\textup{But the roots are non-zero. Therefore k=7}" /></a>