# 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 = 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}**

x^2-kx +7= 0.

One root is 2 times the other.

To solve for x, we assume x1 and 2x1 are the roots of the equation.

We know by the theory of equations that if x1 and x2 are the roots of the equation f(x) = 0, then f(x) = k(x-x1)(x-x2).

Therefore x^2-kx+7 = k(x-x1)(x-2x1), where we choose k =1 so that the coefficients of leading terms x^2 on both sides are equal.

x^2-kx+7 =(x-x1)(x-2x1)

x^2-kx+7 = x^2-(x1+2x1)x+2x1^2

x^2-kx = x^2-3x1+2x1^2

Equating x's and constanterms on both sides we get:

-k = - 3x1

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

7 = 2x1^2.

So x1^2= 7/2.

x1 = sqrt(7/2) or -sqrt(7/2).

So from (1) k = 3x1 = sqrt(7/2) . Or k = -sqrt(7/2).