The roots of the quadratic equation 2x^2 - ax + 3 = 0 (where a >0) are a and b, while those of the equation....?

The roots of the quadratic equation 2x² - ax + 3 = 0 (where a >0) are α and β, while those of the equation 9x² - 52x + 4 = 0 are 1/α² and 1/β². Find value of a.


The roots of the quadratic equation 2x² - ax + 3 = 0 (where a >0) are α and β.
Sum of the roots: α + β = a/2 ...... [1]
Product of the roots: αβ = 3/2 ...... [2]

The roots of the equation 9x² - 52x + 4 = 0 are 1/α² and 1/β².
Sum of the roots:
(1/α²) + (1/β²) = 52/9
(α² + β²) / (αβ)² = 52/9
[(α² + 2αβ+ β²) - 2αβ] / ( αβ)² = 52/9
[(α + β)² - 2αβ] / ( αβ)² = 52/9 ...... [3]

Substiture [1] and [2] into [3] :
[(a/2)² - 2(3/2)] / ( 3/2)² = 52/9
[(a²/4) - 3] / (9/4) = 52/9
[(a²/4) - 3] * (4/9) = 52/9
[(a²/4) - 3] * 4 = 52
a² - 12 = 52
a² - 64 = 0
(a - 8)(a + 8) = 0
a = 8 or a = -8 (rejected, for given that a > 0)

Ans: a = 8

Not possible for one of the roots of 2x^2 - ax + 3 to be a>0.
Hence, question makes no sense.

Something's wrong with the question. Stating a>0 implies that a is real number, but for a to be a root of the first equation then the left side must be zero when evaluated with x=a:

2(a)^2 - a(a) + 3 = 0
2a^2 - a^2 + 3 = 0
a^2 = -3/2

That's not possible for any real number a.