Question

are roots of (b-c) x^2+2(c-a) x+a-b=0 real?​

Answer 1

Answer:

Roots will be equal if and only if Descriminant is equal to 0 i.e b^2 -4ac=0……so in above quadratic equation b can be replaced with (c-a)

and a =(b-c) and c=(a-b)….so putting value of a,b,c in above equation we have…..

(c-a)^2–4(b-c)*(a-b)

=> (c-a)^2=4(b-c)(a-b)

=> c^2+a^2–2ac=4(ab-b^2-ac+cb)

=> c^2+a^2–2ac=4ab-4b^2–4ac+4cb

=> c^2+a^2–2ac +4ac=4ab-4b^2+4cb

=> (c+a)^2+(2b)^2–4(ab+cb)=0

=> (c+a)^2+(2b)^2–2(2b)(a+c)=0

=> (c+a-2b)^2 =0

=> c+a=2b

hence proved..

HOPE you understand and like.

Answer 2

Answer:

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Step-by-step explanation:

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