Question

If the sum of the roots of the quadratic equation ax2+bx+c=0 is equal to the sum of the squares of their reciprocals then prove that 2a squarec=c square b+b square a...

Answer 1

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Answer 2

Answer:

let P and Q are roots of eqn

P + Q = -b/a

PQ = c/a

a/c to question,

P + Q = 1/P² + 1/Q²

-b/a ={ ( P + Q)²-2PQ }/P²Q²

-b/a = { b²/a² -2c/a)/(c/a)²

-b×c²/a³ = b²/a² -2c/a

2c/a = b²/a² + bc²/a³ = b( ab + c²)/a³

2ca² = b²a + bc²

2 = b²a/a²c + bc²/a²c

= b²/ac + bc/a²

b²/ac + bc/a² = 2 ( answer)