SOLUTION: Let α and β be the two roots of a quadratic equation ax^2 + 2b^x + c = 0, where a, b and c are constants. Obtain the quadratic equation whose two roots are λ and

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Question 100208: Let α and β be the two roots of a quadratic equation ax^2 + 2b^x + c = 0, where a, b and c are constants.
Obtain the quadratic equation whose two roots are λ and δ such that λ = α^2 + β^2 and δ = α^2 - β^2.
Found 2 solutions by Nate, Edwin McCravy:
Answer by Nate(3500)   (Show Source): You can put this solution on YOUR website!
α = (-2b + sqrt(4b^2 - 4ac))/2a and β = (-2b - sqrt(4b^2 - 4ac))/2a
.
(x - λ)(x - δ) = 0
.
(x - α^2 - β^2)(x - α^2 + β^2) = 0
.
(x - ((-2b + sqrt(4b^2 - 4ac))/2a)^2 - ((-2b - sqrt(4b^2 - 4ac))/2a)^2)(x - α^2 + β^2) = 0
.
(x - (-2b + sqrt(4b^2 - 4ac))^2/4a^2 - (-2b - sqrt(4b^2 - 4ac))^2/4a^2)(x - α^2 + β^2) = 0
.
(x - (4b^2 - 4b*sqrt(4b^2 - 4ac) + 4b^2 - 4ac)/4a^2 - (4b^2 + 4b*sqrt(4b^2 - 4ac) + 4b^2 - 4ac)/4a^2)(x - α^2 + β^2) = 0
.
(x - (8b^2 - 8b*sqrt(b^2 - ac) - 4ac)/4a^2 - (8b^2 + 8b*sqrt(b^2 - ac) - 4ac)/4a^2)(x - α^2 + β^2) = 0
.
(x - (2b^2 - 2b*sqrt(b^2 - ac) - ac)/a^2 - (2b^2 + 2b*sqrt(b^2 - ac) - ac)/a^2)(x - α^2 + β^2) = 0
.
(x + (-2b^2 + 2b*sqrt(b^2 - ac) + ac)/a^2 + (-2b^2 - 2b*sqrt(b^2 - ac) + ac)/a^2)(x - α^2 + β^2) = 0
.
(x + (-4b^2 + 2ac)/a^2)(x - α^2 + β^2) = 0
.
(x + (-4b^2 + 2ac)/a^2)(x - ((-2b + sqrt(4b^2 - 4ac))/2a)^2 + ((-2b - sqrt(4b^2 - 4ac))/2a)^2) = 0
.
(x + (-4b^2 + 2ac)/a^2)(x - (2b^2 - 2b*sqrt(b^2 - ac) - ac)/a^2 + (2b^2 + 2b*sqrt(b^2 - ac) - ac)/a^2) = 0
.
(x + (-4b^2 + 2ac)/a^2)(x + (-2b^2 + 2b*sqrt(b^2 - ac) + ac)/a^2 + (2b^2 + 2b*sqrt(b^2 - ac) - ac)/a^2) = 0
.
(x + (-4b^2 + 2ac)/a^2)(x + 4b*sqrt(b^2 - ac)/a^2) = 0
.
x^2 + x(-4b^2 + 2ac)/a^2 + x*4b*sqrt(b^2 - ac)/a^2 + (-4b^2 + 2ac)4b*sqrt(b^2 - ac)/a^4 = 0
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
Nate's solution is incorrect.

Let α and β be the two roots of a quadratic equation
ax^2 + 2b^x + c = 0, where a, b and c are constants. 
Obtain the quadratic equation whose two roots are λ and δ
such that λ = α^2 + β^2 and δ = α^2 - β^2.

Make use of this

(x - r1)(x - r2) = 0 has roots r1 and r2

and if you multiply that out by FOIL

x² - r1x - r2x + r1r2 = 0

and factor x out of the middle two terms:

x² - (r1 + r2)x + r1r2 = 0

This quadratic equations has roots r1 and r2

Since we want the roots to be λ and δ, we set

r1 = λ, r2 = δ. Substituting:

x² - (λ + δ)x + λδ = 0

Since λ = α² + β and δ = α² - β².

λ + δ = (α² + β²) + (α² - β²) = α² + β² + α² - β² = 2α²

λδ = (α² + β²)(α² - β²) = α4 - β4

So

x² - (λ + δ)x + λδ = 0

becomes

x² - 2α²x + (α4 - β4) = 0

Edwin
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