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Before attempting to solve these, it is better to find out what α+β and αβ is as we can easily see that we can try and use the Vieta's Formula to solve these.
α+β=2.5
αβ=−0.5
Solution (a):
The equation should be in the form x2−(βα+αβ)x+1
We can rewrite βα+αβ to be αβα2+β2. α2+β2 is not hard to find, as it is(α+β)2−2αβ=7.25. Then we can easily get αβα2+β2=−0.57.25=−14.5
Therefore, (x−βα)(x−αβ)=x2+14.5x+1
Solution (b):
The equation should be in the form x2−(α2β+αβ2)x+α3β3
To find α2β+αβ2, we can factorize it to be αβ(α+β). It's lot more easier now, and we get α2β+αβ2=−1.25
Now we need to find α3β3, which is not so difficult as we can write it as (αβ)3=−0.125
Therefore, (x−α2β)(x−αβ2)=x2+1.25x−0.125
Edited:
The equation for (a) can also be written as 2x2+29x+2 while the equation for (b) can be written as 8x2+10x−1 as it will look better.
@Daniel Lim
–
Nope, if the coefficients of a certain quadratic equation are a,b,c and its roots are α,β, then construct a new equation in terms of a,b,c if the roots of the new equation are
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Top NewestBefore attempting to solve these, it is better to find out what α+β and αβ is as we can easily see that we can try and use the Vieta's Formula to solve these.
α+β=2.5
αβ=−0.5
Solution (a):
The equation should be in the form x2−(βα+αβ)x+1
We can rewrite βα+αβ to be αβα2+β2. α2+β2 is not hard to find, as it is(α+β)2−2αβ=7.25. Then we can easily get αβα2+β2=−0.57.25=−14.5
Therefore, (x−βα)(x−αβ)=x2+14.5x+1
Solution (b):
The equation should be in the form x2−(α2β+αβ2)x+α3β3
To find α2β+αβ2, we can factorize it to be αβ(α+β). It's lot more easier now, and we get α2β+αβ2=−1.25
Now we need to find α3β3, which is not so difficult as we can write it as (αβ)3=−0.125
Therefore, (x−α2β)(x−αβ2)=x2+1.25x−0.125
Edited:
The equation for (a) can also be written as 2x2+29x+2 while the equation for (b) can be written as 8x2+10x−1 as it will look better.
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Great :D
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Thanks
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you could rewrite the solution for (a) to be 2x2+29x+2 & of (b) to be 8x2+10x−1. integers seem better in equations than fractions .
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oh, that's right, I will edit it
Thanks ;D
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Nice solution! By the way, here's a challenge for you: can you generalise this to f(x)=ax2+bx+c?:)
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Like what? f(x)=2x2+29x+2 and f(x)=8x2+10x−1?
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a,b,c and its roots are α,β, then construct a new equation in terms of a,b,c if the roots of the new equation are
Nope, if the coefficients of a certain quadratic equation arei) βα,αβ
ii) α2β,αβ2
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For (a),
f(x)=x2−(ac(a−b)2−a2c)x+1
For (b),
f(x)=x2−(ac×a−b)x+(ac)3
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