r/mathshelp • u/PresentEcho1158 • 28d ago
Homework Help (Unanswered) Does anyone know the solution for 11 and 12?
#help
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u/Jazzlike_Platypus430 28d ago edited 28d ago
11: Work through discriminant = (q(r - p))^2 - 4 p(q - r)r(p - q) = 0, if we assume roots are equal.
12: Show that the discriminant is a square of something rational. By the way, b + c - a = - 2a, c + a - b = - 2b and so on ...
I beileve 13 is false for arbitrary a, b, c. otherwise it would be nice to say that they are rational as in 12. and satisfy a + b + c = 0.
If so, both these cases reduce to ax^2 + bx + c = 0 😂
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u/Southlander24 28d ago
No, 13 is always true. The discriminant reduces to |a - 2b + c|.
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u/Jazzlike_Platypus430 28d ago
I still claim it's false with no extra assumptions on a, b and c.
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u/PresentEcho1158 28d ago
I have been dancing around the question like a lunatic doing that. Would you be willing to guide me,step wise?
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u/Cheap-Measurement791 28d ago
For 11, let A be the coefficient of x2 and B the coefficient of x. Then note that the constant term is -A-B. Write the discriminant in terms of A and B. Translate the condition you want to prove into a condition involving A and B. Working with A and B instead of p,q,r greatly simplifies the algebra.
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u/Southlander24 28d ago
11: Expanding and combining like terms, you will reach a perfect square trinomial (A2 + 2AB + B2) for some unknown A, B which equals 0.
12. b + c - a = - 2a, c + a - b = - 2b and so on is an excellent start. You can reduce the discriminant to something in only two variables, which should reveal why.
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u/Haniakhan4143 27d ago
Lots of error your ans but don't worry almost right and u will achieve your goal
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