Let α and β be the roots of the equation 2x^2 + 1 = 4x without finding the values of α and β find (1/α) + (2/β).?

Let α and β be the roots of the equation

2x^2 + 1 = 4x
2x^2 - 4x + 1 = 0
x^2 -2x + 1/2 = 0

From the quadratic we know that
Sum of roots α+β=2
Product of roots αβ=1/2

αβ=1/2 => 1/(αβ)=2
1/α + 1/β = (α+β)/(αβ)
= 2*2 = 4

1/α + 2/β = 4 + 1/β

2x² + 1 = 4x

2x² - 4x + 1 = 0

Polynomial like: ax² + bx + c, where:
a = 2
b = - 4
c = 1

Δ = b² - 4ac (discriminant)

Δ = (- 4)² - 4.(2 * 1) = 16 - 8 = 8 = 4 * 2

√Δ = 2√2


First case:

α = (- b - √Δ)/2a → 1/α = 2a/(- b - √Δ)

β = (- b + √Δ)/2a → 1/β = 2a/(- b + √Δ) → 2/β = 4a/(- b + √Δ)


= (1/α) + (2/β)

= [2a/(- b - √Δ)] + [4a/(- b + √Δ)]

= [2a.(- b + √Δ) + 4a.(- b - √Δ)] / [(- b - √Δ).(- b + √Δ)]

= 2a.[(- b + √Δ) + 2.(- b - √Δ)] / [b² - Δ]

= 2a.(- b + √Δ - 2b - 2√Δ) / (b² - Δ)

= 2a.(- 3b - √Δ) / (b² - Δ) → recall: Δ = b² - 4ac

= 2a.(- 3b - √Δ) / [b² - b² + 4ac]

= 2a.(- 3b - √Δ)/4ac

= (- 3b - √Δ)/2c → recall: b = - 4 → recall: c = 1 → recall: √Δ = 2√2

= (12 - 2√2)/2

= 6 - √2


Second case:

α = (- b + √Δ)/2a → 1/α = 2a/(- b + √Δ)

β = (- b - √Δ)/2a → 1/β = 2a/(- b - √Δ) → 2/β = 4a/(- b - √Δ)


= (1/α) + (2/β)

= [2a/(- b + √Δ)] + [4a/(- b - √Δ)]

= [2a.(- b - √Δ) + 4a.(- b + √Δ)] / [(- b - √Δ).(- b + √Δ)]

= 2a.[(- b - √Δ) + 2.(- b + √Δ)] / [b² - Δ]

= 2a.(- b - √Δ - 2b + 2√Δ) / (b² - Δ)

= 2a.(- 3b + √Δ) / (b² - Δ) → recall: Δ = b² - 4ac

= 2a.(- 3b + √Δ) / [b² - b² + 4ac]

= 2a.(- 3b + √Δ)/4ac

= (- 3b + √Δ)/2c → recall: b = - 4 → recall: c = 1 → recall: √Δ = 2√2

= (12 + 2√2)/2

= 6 + √2



Conclusion:

(1/α) + (2/β) = 6 ± √2

2x² + 1 = 4x

Re arrange the equation to get. .

2x² - 4x + 1 = 0

a = 2, b = -4, c = 1

α + β = -b/a = -(-4/2) = 4/2 = 2

αβ = c/a = 1/2

Find (1/α) + (1/β)

Find the L.C.M to get. ...

= (β + α)/αβ

= (α + β)/αβ

Remember that: α + β = 2 and αβ = 1/2

= (2)/(1/2)

= 2 ÷ 1/2

= 2 × 2/1

= 4

The question says that without finding α and β

So, you need to find the L.C.M and use sum and product. ... the way I solved it