If $a,b$ are the roots of the equation $2 x^2 -3 x +1 = 0$, find an equation whose roots are $a/(2b +3)$, $b/(2a +3)$
I was practicing quadratic equation questions online but I am stuck on this question.
I found the values of roots $a= 1$, $b= 1/2$
Asked
Active
Viewed 214 times
0
aMighty
- 267
-
Did you find the roots $a$ and $b$? – ploosu2 Mar 12 '15 at 21:12
-
value of roots a= 1, b= 1/2 – aMighty Mar 12 '15 at 21:16
4 Answers
4
The roots have product $\,=\color{#c00}{ab}/(4\color{#c00}{ab}+6(\color{#0a0}{a\!+\!b}) + 9)\,$ and we know $\,\color{#c00}{ab}=1/2,\ \color{#0a0}{a+b}=3/2.\,$ Similary their sum is symmetric in $\,a,b\,$ so it can be written as a polynomial in $\,ab\,$ and $\,a\!+\!b.$
This gives their sum $\,s\,$ and product $\,p.\,$ By Vieta they are the roots of $\,x^2-s\,x+p\, =\, 0.$
Remark $\ $ Note that this way we don't need to compute the roots. Rather, we derive a formula that maps the coefficients of the quadratic for the original roots into the coefficients of the quadratic for the transformed roots. This works universally for any given quadratic - whose roots might be hairy irrational numbers. But this way avoids calculating with irrationals, instead working only with the simpler rational coefficients.
This is a simple example of exploiting the innate symmetry of a problem - here that the coefficients are (elementary) symmetric polynomials of the roots, and every symmetric polynomial can be written as a polynomial in these elementary symmetric polynomials. In fact there is a simple algorithm to do by Gauss (though that would be overkill here).
Bill Dubuque
- 272,048
-
you've edited your post n number of times :P :D and it's worth it :D Good job Thank you for your precious comment God bless you :) – aMighty Mar 12 '15 at 21:37
1
Your new equation is $(x-\frac a{2b+3})(x-\frac b{2a+3})=0$ Do you see why? Can you find $a$ and $b$?
Ross Millikan
- 374,822
0
use the Quadratic root formula to find the values of a and b. calculate the two new roots: $a'=\frac{a}{2b+3}$ and $b'=\frac{b}{2a+3}$. Simply plug these new values within $$(x - a')(x - b')$$ Now multiply the two parentheses and you're done.
rodrigo
- 171
-
-
@amighty No, you're not done because the product still involves $,a,$ and $,b,,$ You don't need to compute the roots if your exploit the innate symmetry - see my answer. – Bill Dubuque Mar 12 '15 at 21:31
-
@BillDubuque I've seen your post and it's good. Thank you for being so helpful :) – aMighty Mar 12 '15 at 21:40
0
I think it is better to not to calculate the roots. The new equation is $x^2-Sx+P=0$ where $$P={a\over 2b+3}.{b\over 2a+3}={ab\over 4ab+6(a+b)+9}={{1\over 2}\over 4\times{1\over 2}+6({3\over 2})+9}$$ and $$S={a\over 2b+3}+{b\over 2a+3}={2a^2+3a+2b^2+3b\over 4ab+6(a+b)+9}={2\big((a+b)^2-2ab)\big)+3(a+b)\over 4ab+6(a+b)+9}$$
Fermat
- 5,230