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Question

# If a, b, c are positive rational numbers such that a > b > c and the quadratic equation (a+bâˆ’2c)x2+(b+câˆ’2a)x+(c+aâˆ’2b)=0 has a root in the interval (-1, 0), then

A

b + c > a

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B

c + a < 2b

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C

both roots of the given equation ae rational

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D

the equaiton ax2+2bx+c=0 has both negative real roots

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Solution

## The correct options are B c + a < 2b C both roots of the given equation ae rational D the equaiton ax2+2bx+c=0 has both negative real roots a>b>c ...(i) and given equation is (a+b−2c)x2+(b+c−2a)x+(c+a−2b)=0 ...(ii) ∵ Eq. (ii) has a root in the interval (-1, 0) ∴ f (-1) f(0) < 0 ⇒(2a−b−c)(c+a−2b)<0 ...(iii) From Eq. (i) a>b⇒a−b>0 and a>c⇒a−c>0∴2a−b−c>0 .... (iv) From Eqs. (iii) and (iv), c+a−2b<0 or c+a<2b. ∴ Alternate (b) is correct Discriminant of Eq. (ii) =(b+c−2a)2−4(a+b−2c)(c+a−2b)={(b−a)+(c−a)}2−4{(a−c)+(b−c)}{c−b+a−b}={(c−a)−(a−b)}2−4{(b−c)−(c−a)}×{(a−b)−(b−c)}If b−c=X,c−a=Y and a−b=Z then, discriminant=(Y−Z)2−4(X−Y)(Z−X)=(2X−Y−Z)2 (∵X+Y+Z=0)=(0+3X)2=9X2=9(b−c)2 ∴ Both roots of the given equation are rational. Alternate (c) is correct. If roots of ax2+2bx+c=0 are α and β Then α+β=−2ba<0∴α+β<0 and αβ=ca>0 (∵a>b>c) and discriminant of ax2+2bx+c=0is 4b2−4ac>0 (∵b>c and b<a) Hence, roots of ax2+2bx+c=0 are negative and real. Alternative (d) is correct

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