determine the values of k so that 1 root is double the other root x^2-kx+7  

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x^2 - kx + 7 = 0

==> a= 1   b= -k    c= 7

Let x1 and x2 be the roots such that:

x2= 2x1

Using viete's rule we know that:

x1+ x2= -b/a = k

==> x1+ 2x1 = k

==> 3x1= k ........(1)

Also:

x1*x2 = c/a =...

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x^2 - kx + 7 = 0

==> a= 1   b= -k    c= 7

Let x1 and x2 be the roots such that:

x2= 2x1

Using viete's rule we know that:

x1+ x2= -b/a = k

==> x1+ 2x1 = k

==> 3x1= k ........(1)

Also:

x1*x2 = c/a = 7

==> x1(2x1) = 7

==> 2x1^2 = 7

==> x1^2 = 7/2

==> x1= +-sqrt(7/2)

But we know that: x2= 2x1

==> x2= +-2sqrt(7/2)

Also, from (1) we know that:

k= 3x1

==> we have 2 possible values for K.

==> k = 3x1= +-3sqrt(7/2)

==> K = { -3sqrt(7/2) ,  3sqrt(7/2}

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