Question

Let ABC be a triangle with AB = BC. Prove that triangle ABC is an obtuse triangle if and only if the equation $Ax^{2} +Bx+ C = 0$ has two distinct real roots. You may assume A;B;C are measured in either degrees or radians.

Answer 1

Given :  ABC be a triangle with AB = BC.

To Find : Prove that  triangle ABC is an obtuse triangle if and only if the equation  has two distinct real roots

Solution:

AB = BC

=> ∠A = ∠C

∠B = π -  A - C    

=> ∠B = π -  2A

Triangle is obtuse if   ∠B > π/2

Ax² + Bx + C = 0

=> Ax²  + (π -  2A)x  + A  =  0

=> x²  + (π -  2A)x/A  + 1  =  0

=> x²  + (π/A -  2)x   + 1  =  0  

case 1 :  if Roots are not real

=>  (π/A -  2)² - 4(1) < 0

=>   (π/A -  2)²  <  4

=> -  2 < (π/A -  2) < 2

=>  - 4 < π/A <  4

=>  π/A <  4

=> π/4 <  A

=> π/2 <  2A

   

=> ∠B = π -  2A < π/2 hence not obtuse

case 2 :  if Roots are  real  and Equal

(π/A -  2)² - 4(1) = 0

=> - 4 = π/A =  4

=>  π/4 =  A

∠B = π -  2A =  π/2 hence not obtuse

case 3 if Roots are  real  and distinct

(π/A -  2)² - 4(1) >  0

(π/A -  2) > 2

=> π/A  > 4

=> A < π/4

Hence ∠B = π -  2A >  π/2 so  obtuse

QED

Hence proved

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