9(x^2+1/x^2)-27(x+1/x)+8=0
Answers
given equation is 9(x^2 + 1/x^2) - 27(x + 1/x) + 8 = 0
let x + 1/x = t
9(t^2 - 2) - 27t + 8 = 0
9t^2 - 27t - 10 =0
(3t - 10)(3t + 1) = 0
t = 10/3 , -1/3
t = 10/3
x + 1/x = 10/3
3x^2 - 10x + 3 = 0
(3x - 1)(x - 3) = 0
x = 3 , 1/3
t = -1/3
x + 1/x = -1/3
3x^2 + x + 3 = 0.............................(1)
D = b^2 - 4ac
= 1 - 36
= -35 (negative)
equation (1) is always positive ,hence no root
root of the given equation is x = 3, 1/3
Answer:
x= 3 and ⅓
The equation will be : 3x² + x + 3.
Step-by-step explanation:
9(x² + 1/x²) - 27(x + 1/x) + 8 = 0
let x + 1/x = t
then,
9(t² - 2) - 27t + 8 = 0
9t² - 27t - 10 =0
(3t - 10)(3t + 1) = 0
t = 10/3 , -1/3
3x²-10x+3= 0
(3x-1)(x-3) = 0
x= 3 , ⅓
x+1/x= -1/3
The equation will be : 3x² + x + 3.
Roots will be :
b²-4ac
= 1 -36
= -35
x= 3 and ⅓
- Case I: b² – 4ac > zero
When a, b, and c are actual numbers, a ≠ zero and the discriminant is high quality, then the roots α and β of the quadratic equation ax² +bx+ c = zero are actual and unequal.
- Case II: b²– 4ac = zero
When a, b, and c are actual numbers, a ≠ zero and the discriminant is zero, then the roots α and β of the quadratic equation ax²+ bx + c = zero are actual and equal.
- Case III: b²– 4ac < zero and ideal square
When a, b, and c are actual numbers, a ≠ zero and the discriminant is high quality and ideal square, then the roots α and β of the quadratic equation ax² + bx + c = zero are actual, rational and unequal.
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