What is the solution to this problem?

Question

Find a quadratic equation width integral coefficients whose roots are thrice those of 3x sq. + 5x - 2 = 0

Answer 1

gudspeling's answer is correct. And it is how to work it out rigorously. But, there is a trick you can follow should you ever need to do this:

Let's say you have: 8x^2 + 5x - 3 = 0 and need to multiply the roots by, say, 8. All you have to do is divide x^2's coefficient by the multiplier (8 in this case) and multiply the numerical term by the multiplier: 8/8 x^2 + 5x - 3*8 = 0 = x^2 + 5x - 24.

Works beautifully, and is much faster than proving your answer with its derivation!

Naturally, knowing how to do it is important! I do not mean to say it isn't. But how many people really remember the derivation of why the first derivative of 3x^7 is 21x^6? This is a little similar to that: just remember the result and you'll almost certainly be A-OK.

Answer 2

The equation is 3x² + 5x -2 = 0. (*)
The trick is to let x = y/3, then y = 3x.
Substututing this into (*) we get
y²/3 + 5y/3 -2 = 0
or, clearing denominators,
y² + 5y -6 = 0.
You can check this by finding the roots of each
equation and comparing them.

Answer 3

3x² + 5x - 2 = 0

Sum of roots: -5/3
Product of roots: -2/3

If the roots are tripled:
Sum of roots = -5/3 * 3 = -5
Product of roots = -2/3 * 3² = -6

-b/a = -5
c/a = -6
If a = 1
b = 5
c = -6

Required quadratic:
x² + 5x - 6 = 0

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Alternatively:
3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x+2)-1(x+2) = 0
(3x-1)(x+2) = 0
x = {1/3, -2}
3x = {1, -6}

New quadratic:
(x-1)(x+6) = 0
x² + 5x - 6 = 0

Answer 4

0...duh.