Maths Olympiad Prep

Library / /156 of 520

Algebra Difficulty 6.3 National olympiad Prove it

4・ 11 When a,b,ca, b, c are real numbers, prove that the equation
x2−(a+b)x+(ab−c2)=0x^{2}-(a+b) x+\left(a b-c^{2}\right)=0

has two real roots, and find the condition for these roots to be equal.

Solution

[Solution] The discriminant of the given equation is
Δ=(a+b)2−4(ab−c2)=(a−b)2+4c2,\Delta=(a+b)^{2}-4\left(a b-c^{2}\right)=(a-b)^{2}+4 c^{2},

Since aa, bb, and cc are all real numbers, hence Δ⩾0\Delta \geqslant 0, so the equation has two real roots. The condition for these two roots to be equal is

i.e., □\square
(a−b)2+4c2=0(a-b)^{2}+4 c^{2}=0
a=b,c=0.a=b, c=0 .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.