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

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We have the equation 4x^2 -21x + 20 = 0.

4x^2 -21x + 20 = 0

=> 4x^2 – 16x – 5x + 20 = 0

=> 4x (x – 4) – 5(x – 4) = 0

=> (4x -5) (x – 4) =0

The roots of this equation are 5/4 and 4.

Now the quadratic with roots twice that of 5/4 and 4 is:

(x – 5/2)(x – 8) = 0

=> x^2 – 5x/2 – 8x + 20 =0

=> x^2 – 21x/2 + 20 =0

=> 2x^2 – 21x + 40 = 0

**The required equation is 2x^2 – 21x + 40 = 0.**

We'll put the quadratic equation that has to be found:

ay^2 + by + c = 0, where y1 = 2x1 and y2 = 2x2

We'll write Viete'e relations for the quadratic that has to be determined:

y1 + y2 = -b/a

2x1 + 2x2 = -b/a

2(x1 + x2) = -b/a

We'll divide by 2:

x1 + x2 = -b/2a (1)

But x1 and x2 are the roots of the given equation:

4x^2 -21x + 20 = 0

x1 + x2 = 21/4 (2)

We'll substitute (2) in (1):

21/4 = -b/2a

We'll simplify by 2:

21/2 = -b/a

We'll cross multiply:

-2b = 21a

y1*y2 = c/a

4x1*x2 = c/a

x1*x2 = c/4a (3)

But x1*x2 = 20/4 = 5 (4)

We'll substitute (4) in (3):

5 = c/4a

c = 20a

Substituting a,b,c, we'll get:

2x^2 – 21x + 40 = 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.