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

#1**0 **

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.

Add 9/16 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.

Add 3/4 to both sides:

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.

Add 3/4 to both sides:

**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**+1 **

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