How do you solve the system #y^2=16x# and #4x-y=-24#?

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Answer 1 · 2016-08-24T17:51:52

#(1) : "There is no Soln. in" RR.#

#(2) : "In" CC, "the soln. is" x=1/2(-11+-isqrt23), y=2+-2isqrt23#.

#4x-y=-24 rArr 4x=y-24#

Sub.ing, in #y^2=16x#, we have, #y^2=4(y-24)=4y-96#, i.e.,

#y^2-4y=-96#

Completing the square on the L.H.S., #y^2-4y+4=4-96=-92#

#rArr (y-2)^2=-92<0# which is not possible in #RR#.

Hence, the System of eqns. have no Soln. in #RR#.

However, in #CC#, we have, #y-2=+-2isqrt23 rArr y=2+-2isqrt23#

Then, from #4x=y-24, "we get", x=1/4(2+-2isqrt23-24)#

#=1/4(-22+-2isqrt23)=1/2(-11+-isqrt23)#.

Enjoy maths.!