Quadratic equations always trouble you!

Algebra Level 3

Select the most appropriate option :

If $a<b<c<d$ and $k$ is a real number independent of $x$ , then $(x-a)(x-c) + k(x-b)(x-d)$ has:

All real roots All integral roots Imaginary roots Without knowing the values of

a,b,c,d

, nothing can be said

1 solution

Asad Jawaid
Nov 3, 2015

by using the discriminate we get {(x-b)(x-d)} ^2.

Through this we can say that all integral roots are real roots.b and d have to positive cause if b was less than d,that would contradict the question.They could be integral roots for what we know,but it has not said that and + non-integral root b can be less than + non-integral root d,so we use this to assume that they are real.

Moderator note:

Can you explain what you mean in "By using the discriminant (edited) we get $[(x-b)(x-d)]^2$ ?
Are you considering this as a quadratic in $x$ ? If so, $x$ should not appear in the discriminant.
Are you considering this as a quadratic in $k$ ? If so, we need a $k^2$ term.

Can you explain what you mean in "By using the discriminant (edited) we get $[(x-b)(x-d)]^2$ ? Are you considering this as a quadratic in $x$ ? If so, $x$ should not appear in the discriminant.
Are you considering this as a quadratic in $k$ ? If so, we need a $k^2$ term.

Calvin Lin Staff - 5 years, 7 months ago

i had the answer but i was not sure of a valid reason,cause i thought of it mentally....so can you please explain it better? and yeah i was considering it as a quadratic. And i meant by discriminent b^2-4ac,or is it called determinate? Well i am still a novice. Please help.

Asad Jawaid - 5 years, 7 months ago

Your solution (as it currently stands) is valid, because that is not the discriminant

We could solve this problem by considering the actual discriminant of the quadratic in $x$ , but that is slightly painful to work through.

Calvin Lin Staff - 5 years, 7 months ago

@Calvin Lin – so can you please show the shorter way,that would be nice.

Asad Jawaid - 5 years, 7 months ago

Quadratic equations always trouble you!

1 solution

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