# Thread: A few algebraic arithmetic questions that I'm stuck on....

1. ## A few algebraic arithmetic questions that I'm stuck on....

Any help would be greatly appreciated I'm stuck on about 7 out of 30 questions - Here are a few:

(i)
Given that $cx = \sqrt{\frac{ax^2 - b}{d}}$ express x in terms of a,b,c & d.

Is this just a rearranging as I get a negative in my sqrt?

(ii)
If $x = p + \sqrt{q}$ where p and q are rational, show that $x^2$ and $$x^3/math] are of the form $P + Q\sqrt{q}$ where P and Q are rational. I'm stuck on this one completely (iii) Verify that $z = (4+\sqrt{15})^\frac{1}{3} + (4-\sqrt{15})^\frac{1}{3}$ satisfies $z^3 -3z +8 =0$ Not sure what is ment here... (iv) Given that $\alpha$ and $\beta$ are roots of the equation $x^2 + 3x -6 =0$ find a quadratic equation with integer coefficiants whose roots are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$ I'm stuck with this and cannot find any examples of this online or in my text books. Positive is that I've gotten to grips with LaTex Thanks, Dojo 2. For (i), the first thing you'll want to do is square both sides of the equation. Then you need to isolate x on one side of the equation. It pretty much is a rearranging. For (ii), just multiply out [tex](p + \sqrt{q})(p + \sqrt{q})$$ and note that a product or sum of rational numbers is rational.

3. Originally Posted by dojo

(iii)
Verify that $z = (4+\sqrt{15})^\frac{1}{3} + (4-\sqrt{15})^\frac{1}{3}$ satisfies $z^3 -3z +8 =0$

Not sure what is ment here...
This means if you expand

$\displaystyle \left( (4+\sqrt{15})^\frac{1}{3} + (4-\sqrt{15})^\frac{1}{3}\right)^3 -3\left( (4+\sqrt{15})^\frac{1}{3} + (4-\sqrt{15})^\frac{1}{3}\right) +8$

then simplify you should get zero.

Or just solve $z^3 -3z +8 =0$

What are the solutions??