What is the quadratic equation that has roots twice in magnitude of the roots of 4x^2 -21x + 20 = 0

The given is equation 4x^2-21x+20 = 0. To determine the
quadratic equation whose roots the double the roots of the given
equation.

For the quadratic  equation ax^r+bx+c = 0, the relation
between the roots x1 and x2  is given by:

 Sum of the roots x1+x2 =
-b/a and product of roots x1x2 = c/a.

In the given case , a= 4, b=
-21 and c=20.

Then x1+x2 = -(-21)/4 =
21/4.

x1x2 = 20/4 = 5.

Now let  2x1 and
2x2 be the roots of  a quadratic equation.

Then the sum of the roots
2(x1+x2) = 2(21/4) = 21/2.

Product of the roots = (2x1*2x2) = 4x1*x2
= 4(5)= 20.

Therefore the required equation which has the roots
double of the given equation is :

x^2 - (21/2)x + 20 =
0.

We multiply by 2 to get the integral
coefficients.

2x^2-21x+ 40 = 0.