Question 873184
Complete the Square for {{{y=x^2-Mx+3}}} and expect vertex to have y=0.


{{{y=x^2-Mx+(M/2)^2+3-(M/2)^2}}}, using {{{(M/2)^2}}} to complete the square.
{{{y=(x-M/2)^2+(3-(M/2)^2)}}}


This shows vertex to be {{{x=M/2}}} and {{{highlight_green(y=3-(M/2)^2)}}}
That is what must be zero: the y value.  Solve for M in {{{3-(M/2)^2=0}}};
{{{(M/2)^2=3}}}
{{{M^2=3*4}}}
{{{M^2=3*2*2}}}
{{{M=-2sqrt(3)}}} or {{{M=2sqrt(3)}}}
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FINDING THE POSSIBLE VERTEX INCLUDING x:
The corresponding x values and ordered pair vertices are then
(-sqrt(3), 0)
OR
(sqrt(3), 0)
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Finishing this using the negative valued vertex to demonstrate,
{{{y=(x+sqrt(3))^2+3-(M/2)^2}}}
{{{y=(x+sqrt(3))^2+3-(-2sqrt(3)/2)^2}}}
{{{y=(x+sqrt(3))^2+3-3}}}
{{{highlight(y=(x+sqrt(3))^2)}}}
You can carry through similar steps to use the positive-valued vertex if you want.


{{{graph(300,300,-7,4,-4,7,(x+sqrt(3))^2)}}}