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If the sum of the roots of the quadratic equation ax^2+ bx+c=0. is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in ....
Dear bhanu,
1 Given equation is ax2 + bx + c = 02 Let α, β be the roots. 3 α + β = -ba and αβ = ca [Sum of the roots = -ba and Product of the roots = ca.] 4 α + β = 1 α2+ 1 β2 [Given condition.]5 α + β= α2 + β2 α2 β2 [Simplify.]6 α + β = (α + β)² - 2αβα² β² [Use α2 + β2 = (α + β)2 - 2α β.] 7 -ba =b²c² -2cac²a² [Substitute the values.]8 -ba =b² - 2acc² [Multiply and divide by a2 and simplify.]9 -ba =b²c²-2acc² [Simplify.]10 2acc²=b²c²+ba [Simplify.]11 2a2c = ab2 + bc2 [Simplify.]12 2ab=bc+ca [Divide both sides by abc.]13 Therefore ab, bc, ca are in A.P
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Askiitians Expert
Sagar Singh
B.Tech, IIT Delhi
let A,B be the roots of this equation then,
A+B=-b/a & AB=c/a .................1
now it is given that sum of roots is equal to square of resiprocal of roots
so
A+B=1/A2 +1/B2
A+B=A2+B2 /(AB)2
A+B = [(A+B)2 -2AB]/(AB)2
now putting value of A+B and AB from eq 1
-b/a = [b2/a2 -2c/a] /c2/a2
ab2 +bc2 =2a2c ........................2
now let a/c,b/a,c/b are in HP then
a/b = [b/c + c/a]/2
ab2 + bc2 =2a2c
this result is similar to eq 2 so these terms are in HP.........
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