The quadratic equation whose roots are \(\displaystyle{\sec{{2}}}\theta\)

Zane Decker

Zane Decker

Answered question


The quadratic equation whose roots are sec2θ and cscθ can be
a) x22x+2=0
b) x25x+5=0
c) x27x+7=0
d) x29x+9=0

Answer & Explanation

Kendall Clark

Kendall Clark

Beginner2022-04-09Added 8 answers

Step 1
As others have pointed out, you’ve assume that if uv then u2v2
This is true if u,v are both non-negative, but if both are non-positive, then the inequality reverses, u2v2, and there is nothing we can say if u is positive and v negative.
Another way to write the equation is as
So the monic polynomial with these two roots has
This is really the same as your first approach, but switches to sin and cos immediately.
Kendall Wilkinson

Kendall Wilkinson

Beginner2022-04-10Added 17 answers

Step 1
Recall that
1sec2θ+1csc2θ=1. If
So you want 1x1+1x2=1 You have
p±p2+4p2={1±3if p=2,if p=5,if p=7,if p=9.
When p=2, one of the solutions is negative and so cannot be sec2θ or csc2θ if θ is real. The other one is negative and so cannot be sec2θ or csc2θ if θ is real.
Go on from there is a similar way.

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