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.!

How do you solve the system y^2=16xy^2=16x and 4x-y=-244x-y=-24?