Roots of quadratic equation help URGENT?

Question

The roots of the quadratic equation 16x^2+7x+4=0 are
α² and β².
Find
(i) a quadratic equation whose roots are 1/α² and 1/β².
(ii) two distinct quadratic equations whose roots are α and β.

workings needed thanks

Answer 1

16x² + 7x + 4 = 0
x² + 7/16x + 1/4 = 0
α² + β² = -7/16
α²β² = 1/4

(i)
f(x) = (x - 1/α²) (x - 1/β²)
= x² - (1/α² + 1/β²) x + (1/α²) * (1/β²)
= x² - (β²+α²) / (α²β²) x + 1/(α²β²)
= x² - (-7/16) / (1/4) x + 1/(1/4)
= x² + 7/64 x + 4

(ii)
f(x) = (x - α) (x - β)
= x² - (α + β) x + αβ
= x² - root(α² + 2αβ + β²) x + root(α²β²)
= x² - root(α² + β² + 2root(α²β²)) x + root(1/4)
= x² - root(-7/16 + root(1/4)) x + root(1/4)
At this point, you can choose either the positive or negative for both roots (you have to choose the same sign for both root(1/4)s though). This makes sense since with 2 possible values for each α and β, you would have 4 different equations. Since, α² + β² = -7/16, we know α and β won't both be real, but choosing root(1/4) as 1/2 makes the coefficients real, at least.

Answer 2

The answer to (i) is

16/(x^2) + 7/x + 4 = 0 because then 1/a^2 with give the same result as a^2 in the previous equation. And to get the quadratic in the regular form multiply both sides by x^2 to get

16 + 7x + 4x^2 = 0

Similarly for (ii) the equation 16x^4 + 7x^2 + 4 = 0 would give a zeros for α and β. I believe that the answer to the question would be found by solving this quadratic equation for x^2 using the quadratic formula. The two forms are the +/- outside the square root part.

Answer 3

16/x^2 + 7/x+ 4 = 0

The roots of 16x^2+7x+4=0 are imaginary:
x = [-7 +/-sqrt(7^2- 4*16*4)]/(2*16)
x = -7/32 +/- (3i/32)sqrt(23)
Now Let α² = one of the above roots and solve for α.
Then let β² = the other root and solve for β.
Then form the equation using the factor form of the roots.