ROOTS of quadratic equation help?

Question

The roots of the quadratic equation 16x^2+7x+4=0 are
α²+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.

I meant α² AND β²..

Answer 1

This is a great problem if there's an elegant solution. I'll keep an eye out to see if somebody finds one. In the meantime, here's an ugly brute-force solution.

First of all, I need clarification. Is α²+β² one root, or are α² and β² the two roots? My answer assumes the second option since the question suggests that we are being given both roots. Now continuing on.

It's easy to find the two complex roots by the quadratic equation. Set one equal to α² and the other equal to β².

(i) They don't say that the polynomial must have real coefficients, so let p(x) = (x - (1/α²))*(x - (1/β²)). That's a quadratic with roots 1/α² and 1/β².

(ii) Let q(x) = (x-α)(x-β). Then q(x) and 2*q(x) are two distinct quadratics with roots α and β.

Yes it's a cheap solution, but it fits all the criteria as specified in the problem.

Answer 2

Maybe you really mean: The roots of the quadratic equation 16x^2+7x+4=0 are
α² and β². Then the equation 16(1/x)^2+7(1/x)+4=0 has the roots 1/α² and 1/β². This equation can be rearranged into the quadratic 4x^2+7x+16=0.

Answer 3

Using a and b instead of alpha and beta:

(i)
a^2 + b^2
= (a + b)^2 - 2ab
= (-7/16)^2 - 1/2
= (49-128)/256
= -79/256
ab = 4/16 = 1/4

1/a^2 + 1/b^2
= (a^2 + b^2) / a^2 b^2
= 16( -79/256 )
= -79 / 16
1/a^2 b^2 = 16

The equation is therefore:
x^2 + (79/16)x - 16 = 0
16x^2 + 79x - 256 = 0.

(ii)
I cannot see how to do this part. For equations with real coefficients, the roots have to be in conjugate pairs. I also have no idea how, without solving the original quadratic, to find an equation of which the roots are not symmetrical in relation to a and b.

Answer 4

I put this same equation into my graphing calculator and traced the graph to find the roots. No where in the graph does this line cross the x-axis. Which means that this equation has no roots at all. Which means that 16x^2+7x+4 will never equal 0. No real solution.

Answer 5

16x² + 7x + 4 = 0

Solve using quadratic formula

x = (-b ± √(b² - 4ac))/2a

x = (-7 ± √(49- 256))/32

x = (-7 ± √(-207))/32

x = (-7 ± 3i√(23))/32
.