If a, b, c E R, show that the roots of the equation(a-b) xsq+ (b+c-a)x-c=0 are rational
Answers
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..
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Step-by-step explanation:
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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..