Find the sum of the real values of x satisfying
2x2 + 20x + 21 2x2 + 24x + 11
-------------- = ---------------
2x2 − 12x + 20 2x2 − 8x + 10
The quartic and cubic terms cancel out, leaving a quadratic equation.
(2x^2 + 20x + 21)(2x^2 − 8x + 10) =
(2x^2 + 24x + 11)(2x^2 − 12x + 20)
24x^3 - 98x^2 - 32x + 210 =
24x^3 - 226x^2 - 348x + 220
128x^2 - 316x - 10 = 0
x = {316 ± √(99856 + 5120)} / 256
x = {316 ± √(104976)} / 256
x = {316 ± 324} / 256
x = {-8/256, 640/256}
x = {-1/32, 2.5}
Sum of these values is 79/32 = 2.46875
But if I had remembered that all we are asked for is the sum, I could have just calculated 316/128 = 2.46875
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Posted by Larry
on 2025-04-11 11:44:09 |
Algebraic Solution
