Question

The value of a such that [tex] x^{2} -11 x -p=0 and x^{2} +2px -12=0 may have a common root is

Answer 1

p = -4.708421854  (approximately)
common root:  x =  10.55386  (approximately)

$x^2-11x-p=0\ \ and\ \ x^2+2px-12=0,\ \ \ subtracting,\\\\(2p+11)x=12-p, \ \ \ x=(12-p)/(2p+11)\\\\Multiply\ eq1\ by\ 2p\ ,\ multiply\ eq2\ by\ 11, and\ add\\\\(2p+11)x^2=2p^2+132\\\\x^2=(2p^2+132)/(2p+11)\\\\So,\ \ (12-p)^2=(2p+11)(2p^2+132),\ \ \ simplifing, we\ get\\\\4p^3+288p+21p^2+1308=0$

Substitute p+21/(3*4)=y ie., p = y-7/4 and simplify to get remove the x^2 term.

$4y^3+251.5y+146.875=0\\\\y^3+62.875y+36.71875=0\\\\y^3+3Ay-B=0,\ \ A=62.875/3,\ B=-36.71875\\\\substitute\ y=s-A/s,\ \ t=A/s,\ \ \ you\ get\ s^6-Bs^3-A^3=0\\\\s^3=(B+-\sqrt{B^2+4A^3})/2\\\\Find\ s.\ then\ t=A/s,\ \ \ y=s-t\\\\then\ find\ p=y-7/4\\\\then\ x=common\ root=(12-p)/(2p+11)$

we find
p = -4.708421854  (approximately)
common root:  x =  10.55386  (approximately)