Question

If the sum of the root of the quadratic equation X square + b x square + C is equal to zero is equal to the sum of square of their reciprocal then prove that twice CA square is equal to BC square + a b square

Answer 1

Step-by-step explanation:

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)

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)

Further you can solve it...