Solving these quadratic equations

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chwala
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Homework Statement:

if the roots of the equation ##x^2+bx+c=0## are ##∝, β## and the roots of the equation ##x^2+λbx+λ^2c=0## are ##δ, ϒ##show that the equation whose roots are ##∝ϒ+βδ## and ##∝δ+βϒ## is ##x^2-λb^2x+2λ^2c(b^2-2c)=0##

Relevant Equations:

quadratic equations; sum and roots
1594177309176.png

my original working, i would appreciate alternative methods...
 

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chwala
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pasmith
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If [itex](x - \alpha)(x - \beta) = x^2 + bx + c[/itex] then [itex](x - \lambda\alpha)(x - \lambda\beta) = x^2 + \lambda b + \lambda^2 c[/itex]. So [itex]\{\gamma, \delta\} = \{\lambda\alpha, \lambda\beta\}[/itex].

Now substituting [itex]\gamma = \lambda \alpha[/itex] and [itex]\delta = \lambda \beta[/itex] we find that the roots of the new quadratic are then [tex]
r_1 = \alpha \gamma + \beta \delta = \lambda (\alpha^2 + \beta^2) = \lambda((\alpha + \beta)^2 - 2\alpha\beta) = \lambda(b^2 - 2c)[/tex] and [tex]
r_2 = \alpha \delta + \beta\gamma = 2\lambda \alpha \beta = 2 \lambda c[/tex] yielding [tex]
r_1 + r_2 = \lambda b^2, \qquad r_1r_2 = 2\lambda^2 c (b^2 - 2c)[/tex] as required. Swapping the definitions of [itex]\gamma[/itex] and [itex]\delta[/itex] swaps [itex]r_1[/itex] and [itex]r_2[/itex], leaving [itex]r_1 + r_2[/itex] and [itex]r_1r_2[/itex] unchanged.
 
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PeroK
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i just had a look at his method...i don't really like it...i respect his approach nevertheless...though i prefer old school way of solving quadratics :cool:
The thing to notice here is that it is better to focus on the equations relating roots and coefficents and leave the quadratic equations themselves behind.

1) You have an equation with coefficients ##b, c## and roots ##\alpha, \beta##. This gives:
$$\alpha + \beta = -b, \ \ \alpha \beta = c$$
2) You then have a second equation with coefficents ##b_2, c_2##, say, and roots ##\gamma, \delta##. Hence:
$$\gamma + \delta = -b_2, \ \ \gamma \delta = c_2$$
3) We want a third equation with coefficients ##B, C##, say, and roots ##\alpha \gamma + \beta \delta, \ \alpha \delta + \beta \gamma##, hence:
$$-B = (\alpha \gamma + \beta \delta) + (\alpha \delta + \beta \gamma) = (\alpha + \beta)(\gamma + \delta) = (-b)(-b_2) = bb_2$$ $$C = (\alpha \gamma + \beta \delta)(\alpha \delta + \beta \gamma) = (\alpha^2 + \beta^2)\gamma \delta + (\gamma^2 + \delta^2)\alpha \beta = (\alpha^2 + \beta^2)c_2 + (\gamma^2 + \delta^2)c$$
4) Now you use the simplification:
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = b^2 - 2c \ \ \text{and} \ \ \gamma^2 + \delta^2 = b_2^2 - 2c_2$$
5) This gives us a more general solution to a more general problem:
$$B = - bb_2, \ \ C = (b^2 - 2c)c_2 + (b_2^2 - 2c_2)c$$
6) In this particular case, we have ##b_2 = \lambda b## and ##c_2 = \lambda^2 c##. You can plug those in and get the solution in this case. But, if you specify any other coefficients for the second equation, we already for the solution for that as well.

You see how it was in many ways simpler to break free of the quadratic equations themselves and focus on the algebraic relationships between coefficients and roots. This allows us to solve problems even more generally.
 
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