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The quadratic \(2x^2-3x+27\) has two imaginary roots. What is the sum of the squares of these roots? Express your answer as a decimal rounded to the nearest hundredth.

tertre  Dec 30, 2017
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#1
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Solve for x:

2 x^2 - 3 x + 27 = 0

Write the quadratic equation in standard form.

Divide both sides by 2:

x^2 - (3 x)/2 + 27/2 = 0

Solve the quadratic equation by completing the square.

Subtract 27/2 from both sides:

x^2 - (3 x)/2 = -27/2

Take one half of the coefficient of x and square it, then add it to both sides.

x^2 - (3 x)/2 + 9/16 = -207/16

Factor the left hand side.

Write the left hand side as a square:

(x - 3/4)^2 = -207/16

Eliminate the exponent on the left hand side.

Take the square root of both sides:

x - 3/4 = 1/4 (3 i) sqrt(23) or x - 3/4 = 1/4 (-3 i) sqrt(23)

Look at the first equation: Solve for x.

x = 3/4 + 1/4 (3 i) sqrt(23) or x - 3/4 = 1/4 (-3 i) sqrt(23)

Look at the second equation: Solve for x.

x = 3/4 + 1/4 (3 i) sqrt(23)             or                x = 3/4 - 1/4 (3 i) sqrt(23)

And the sum of their squares = - 99/4

Guest Dec 30, 2017
edited by Guest  Dec 30, 2017
#2
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2x^2 -  3x  +  27  = 0   complete the square

2(x^2 -  3/2x  +  9/16)  =  -27  + 9/8

2(x - 3/4)^2  =  ( -216 + 9) / 8

(x - 3/4)^2  =  -207 / 16

(x - 3/4)^2  =  -207/16

x - 3/4    =  ± 3i√23/4

x =  [ 3  ± 3i√23 ]  3

So.....the roots are

[  3  + 3i√23 ] /  4        and  [ 3  -  3i√23] / 4

And the sum of the squares of these roots  is

[ 9  - 9*23  + 9 - 9*23  ]  /  16   =

(9/16) [ 2  - 46]  =

(9/16) [ -44]  =

-(9/16) (4 * 11)  =

-(99/4)  =

-24.75

CPhill  Dec 30, 2017

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