If x²-4x+1=0 prove that x⁸-194x⁴+1=0
Answers
Answer:
x=2½
Step-by-step explanation:
x²=2 ½*2½ - 4x+1 = 0
x⁸=2½*2½*2½*2½*2½*2½*2½*2½=194*194*194*194=0
Question:
If x² - 4x + 1 = 0 prove that x⁸ - 194x⁴ + 1 = 0.
Step-by-step explanation:
So we have x² - 4x + 1 = 0 and asked to prove that x⁸ - 194x⁴ + 1 = 0. The basic thing that could be done here is to solve the given equation i.e. x² - 4x + 1 = 0 by the quadratic formula (x = [-b ± √(b² - 4ac)/2a].
→ x = [4 ± √((4)² - 4(1)(1))/2(1)]
→ x = [4 ± √(16 - 4)/2]
→ x = [4 ± √(12)/2]
→ x = 4/2 ± 2√3/2
→ x = 2 ± √3
Now, substitute the value of x in x⁸ - 194x⁴ + 1 = 0.
If x = 2 + √3
→ (2 + √3)⁸ - 194(2 + √3)⁴ + 1 = 0
→ 37633.99 - 194(193.99) + 1 = 0
→ 37633.99 - 37634.06 + 1 = 0
→ 1 ≠ 0
If x = 2 - √3
→ (2 - √3)⁸ - 194(2 - √3)⁴ + 1 = 0
→ 2.66 - 194(0.005) + 1 = 0
→ 2.66 - 0.97 + 1 = 0
→ 2.69 ≠ 0
So, it is impossible to prove that x⁸ - 194x⁴ + 1 = 0 by quadratic formula. Since, we have no direct value of x⁸ or x⁴. So, let's try to solve this problem as: x⁴ + 1/x⁴.
→ x² - 4x + 1 = 0 (given equation)
Take the negative term on R.H.S.
→ x² + 1 = 4
Divide by x on both sides,
→ x²/x + 1/x = 4/x
→ x + 1/x = 4
Do squaring on both sides,
→ (x + 1/x)² = (4)²
→ x² + 1/x² + 2 = 16
Used identity: (a + b)² = a² + b² + 2ab
→ x² + 1/x² = 14
Square up both sides again, using identity (a + b)² we get this following result,
→ x⁴ + 1/x⁴ = 194
→ x⁴ = 194 - 1/x⁴
Take x⁴ common from x⁸ - 194x⁴ + 1 = 0.
→ x⁴(x⁴ - 194) + 1 = 0
Substitute the value of x⁴ in the above equation,
→ (194 - 1/x⁴)(x⁴ - 194) + 1 = 0
→ 194(x⁴ - 194) - 1/x⁴(x⁴ - 194) + 1 = 0
→ 194x⁴ - 37636 - 1 + 194/x⁴ + 1 = 0
→ 194x⁴ + 194/x⁴ - 37636 = 0
194x⁴ + 194/x⁴ can be written as 194(x⁴ + 1/x⁴) and from above we have value of x⁴ + 1/x⁴ is 194. So,
→ 194(194) - 36736 = 0
→ 37636 - 37636 = 0
→ 0 = 0
Hence, proved.