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

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### 2 Answers

You may evaluate the solutions of the given quadratic equation, using quadratic formula, such that:

`x_(1,2) = (-(-21) +- sqrt((-21)^2 - 4*4*20))/(2*4)`

`x_(1,2) = (21 +- sqrt(441 - 320))/8`

`x_(1,2) = (21 +- sqrt(121))/8 => x_(1,2) = (21 +- 11)/8`

`x_1 = 4 ; x_2 = 5/4`

The problem provides the information that the solution to quadratic equation you need to find are double the solutions `x_1 = 4 ; x_2 = 5/4` , such that:

`x_3 = 2*4 = 8 `

`x_4 = 2*5/4 => x_4 = 5/2`

You may write the factored form of quadratic equation, such that:

`a(x - x_3)(x - x_4) = a(x - 8)(x - 5/2)`

**Hence, evaluating the quadratic equation, under the given conditions, yields **`(x - 8)(x - 5/2).`

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