Asked by simran
solve by completing square method 16x^2-24x-13=0
Answers
Answered by
Steve
16x^2-24x-13=0
16(x^2 - 3/2 x) = 13
16(x^2 - 3/2 x + (3/4)^2 = 13+16(3/4)^2
16(x-3/4)^2 = 22
(x-3/4)^2 = 11/8
x = 3/4 ±√(11/8)
or, x = (3±√22)/4
or, eschewing fractions,
16x^2-24x-13 = 0
16x^2-24x+9 = 13+9
(4x-3)^2 = 22
...
16(x^2 - 3/2 x) = 13
16(x^2 - 3/2 x + (3/4)^2 = 13+16(3/4)^2
16(x-3/4)^2 = 22
(x-3/4)^2 = 11/8
x = 3/4 ±√(11/8)
or, x = (3±√22)/4
or, eschewing fractions,
16x^2-24x-13 = 0
16x^2-24x+9 = 13+9
(4x-3)^2 = 22
...
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