I came across this question in Cambridge. I tried for ages, and I couldn't get it.
For the equation x^2 = 5x - 8, find the value of a^(1/3) + b^(1/3), where a and b are the roots of the equation, without solving the equation.
Thanks.
I came across this question in Cambridge. I tried for ages, and I couldn't get it.
For the equation x^2 = 5x - 8, find the value of a^(1/3) + b^(1/3), where a and b are the roots of the equation, without solving the equation.
Thanks.
Mainly because the question says you're not meant to. But if you do solve it, and find values for a and b you'll end up with complex roots and it doesn't look pretty when you take the cube root of that. Essentially you're meant to have the final answer as -1 or something with a root 21 in it. I'll check and post the answers up when I get home.
The equation is $\displaystyle x^2- 5x+ 8= 0$. If a and b are roots of that equation, then $\displaystyle (x- a)(x- b)= x^2- (a+ b)x+ ab= x^2- 5x+ 8$ so we must have a+ b= 5 and ab= 8. Now we can use $\displaystyle a+ b= (a^{1/3}+ b^{1/3})(a^{2/3}+ (ab)^{1/3}+ b^{2/3})$ which, with a+ b= 5, ab= 8 becomes $\displaystyle 5= (a^{1/3}+ b^{1/3})(2+ a^{2/3}+ b^{2/3})$.
Now, look at the fact that [tex](a^{1/3}+ b^{1/3})^2= a^{2/3}+ 2(ab)^{1/3}+ b^{2/3}= 4+ a^{2/3}+ b^{2/3}. That is, $\displaystyle 2+ a^{2/3}+ b^{2/3}= -2+ (a^{1/3}+ b^{1/3})^2$. Putting that into the previous equation, and letting $\displaystyle a^{1/3}+ b^{1/3}= X$ we have $\displaystyle 5= X(-2+ X^2)$ or $\displaystyle X^3- 2X- 5= 0$.
We are still not to the bottom of this question.
Sorry HallsofIvy, but you lost a negative sign in your factorisation of $\displaystyle a+b,$ the middle term in the second bracket should be negative.
Following that through I think that the equation at the end should be $\displaystyle X^{3}-6X-5=0.$
This has an obvious solution of $\displaystyle X=-1,$ and the other two are then found to be $\displaystyle 2.7913$ and $\displaystyle -1.7913$ (4dp).
I suppose that it is the positive root that we are looking for.