anonymous
  • anonymous
The roots of 2x^2-3x=4 are a and b. Find the simplest quadratic equation which has roots 1/a and 1/b
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
replace x with 1/x and simplify
anonymous
  • anonymous
You're an angel from heaven. Just another question. If instead of 1/a and 1/b, it were to be (b+1/a) and (a+1/b), what would you do?
DebbieG
  • DebbieG
So you want: the simplest quadratic equation which has roots (b+1/a) and (a+1/b), where a and b are the roots of the original equation given above, right? Can you just solve the equation above to find your a and b (ugly irrational expressions, lol), and then use those to cook up: m= (b+1/a) n=(a+1/b) Which will be REALLY ugly expressions. :) Then you cook up: y=(x-m)(x-n) You might want to multiply by a constant: y=a(x-m)(x-n) to rid yourself of fractional coefficients... depends on how you define "simplest".

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ganeshie8
  • ganeshie8
maybe just add both the eqations :)
DebbieG
  • DebbieG
This is a fascinating problem.... I was playing around with the original question, because my initial thought was to take a different approach than @ganeshie8 did, and my solutions looked different... but I have convinced myself that they are equivalent, lol.
ganeshie8
  • ganeshie8
2x^2-3x=4 has roots a, b you knw another equaiton whose roots are 1/a, 1/b now, think wat we need to do, to find an equation whose roots are a+1/a, b+1/b
ganeshie8
  • ganeshie8
oh you have mixed them, u wanto find a quadratic whose roots are a+1/b, b+1/a
ganeshie8
  • ganeshie8
@DebbieG thats the straightforward way of doing this.. (bruteforce :) )
anonymous
  • anonymous
for our easiness we consider that the roots of 2x^2-3x-4=0 are c and d. and we consider that 2=a and -3=b and -4=c, we know that c+d=-b/a and cd=c/a, so c+d=-(-3)/2 c+d=3/2 now cd=c/a cd=-4/2 cd=-2. now we have to form equation whose roots are 1/c and 1/d . now (1/c)+(1/d)=(c+d)/cd putting values of c+d and cd \[(c+d)/cd=(3\div2)\div(-2)\]\[(c+d)/cd= -3/4\] now (1/c)(1/d)=1/cd now 1/cd=1/-2 1/cd=-1/2 \[1/cd=-1/2\] now we know that x^2-(sum of root )x+(product of root)=0 S.O.R=-3/4 and P.O.R=-1/2 now x^2-(-3/4)x+(-1/2)=0 x^2+3x/4-1/2=0 (4x^2+3x-2)/4=0 4x^2+3x-2=0 \[4x ^{2}+3x-2=0\] answeer
anonymous
  • anonymous
Ibad - If instead of 1/a and 1/b, it were to be (b+1/a) and (a+1/b), what would you do?
anonymous
  • anonymous
simply we will add it and our objective would be to form an expression of the new roots which contains old roots in form a+b and ab, to make a new equation of new roots, (b+1)/a + (a+1)/b {b(b+1)+a(a+1)}/ab (b^2+b+a^2+a)/ab (a+b+a^2+b^2)/ab {a+b+((a+b)^2)-2ab}/ab now we have expression in a+b or ab form now we will put the value of a+b and ab from old equation and we will get sum of roots same we will do for product of roots (b+1)/a x (a+1)/b {(b+1)(a+1)}/ab {ab+b+a+1}/ab (ab+a+b+1)/ab now this expression is also in ab and a+b form ow we will put the value of a+b and ab from old equation and we will get product of roots and yeah new equation will be in this form x^2-(sum of root)x+(Product of root)=0
DebbieG
  • DebbieG
@prasidharora, you said you needed roots \[\large b+\frac{ 1 }{a }~and~a+\frac{ 1 }{b}\] Is that what you means? Or do you mean \[\large \frac{ b+1 }{a }~and~\frac{ a+1 }{b}\]
DebbieG
  • DebbieG
*meant
anonymous
  • anonymous
I meant the one on top.
DebbieG
  • DebbieG
OK, that's what I thought. I really like @ibad's approach on the first question, so let me see if I can apply that here as well. From the polynomial that we started with, we had roots of a & b. So we can say: \(\Large 2(x^2-\dfrac{3}{2}x-2)=0\) has roots a & b, and so: \(\Large x^2-\dfrac{3}{2}x-2=(x-a)(x-b)=x^2-x(a+b)+ab\) Therefore: \(\Large a+b=\dfrac{3}{2}\) and \(\Large ab=-2\) Now save those, we'll come back to use them later. Now we want to cook up a polynomial with the roots: \(\large m=b+\dfrac{ 1 }{a }~and~n=a+\dfrac{ 1 }{b}\) so that polynomial has, as factors, (x-m) and (x-n) So we get: \(\Large (x-m)(x-n)=x^2-x(m+n)+mn\) so using \(\large m=b+\dfrac{ 1 }{a }~and~n=a+\dfrac{ 1 }{b}\) we get: \(\large m+n=b+\dfrac{ 1 }{a }+a+\dfrac{ 1 }{b}=\dfrac{ 2ab+(b+a) }{ ab }\) and \(\large mn=(b+\dfrac{ 1 }{a })(a+\dfrac{ 1 }{b})=ab+2+\dfrac{1}{ab}=\dfrac{ (ab)^2+2ab+1 }{ ab }\) OK now to put it all together: \(\large m+n=\dfrac{ 2ab+(b+a) }{ ab }=\dfrac{ 2(-2)+\dfrac{3}{2} }{ -2 }=2-\dfrac{3}{4}=\dfrac{5}{4}\) and \(\large mn=\dfrac{ (ab)^2+2ab+1 }{ ab }=\dfrac{ 4+2(-2)+1 }{ -2 }=-2+2-\dfrac{1 }{ 2 }=-\dfrac{1 }{ 2 }\) So putting that into \(\Large x^2-x(m+n)+mn\) yields: \(\Large x^2-x(\dfrac{5}{4})-\dfrac{1 }{ 2 }\) So we can get: \(\Large y=4\cdot[x^2-x(\dfrac{5}{4})-\dfrac{1 }{ 2 }]\) \(\Large y=4x^2-5x-2\) OK, I'm trying to confirm that this function has the right roots (related to the roots of the original equation), using Wolfram, and it is NOT WORKING so I guess I have an error somewhere. :( But I think the method is sound, I'll look at it and see if I can figure out why it isn't working.

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