+4609 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.
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
+128408 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
