If x be real prove that x 2 + 34 x - 71 x 2 + 2 x - 7 can have no value between 5 and 9 .

Let x 2 + 34 x - 71 x 2 + 2 x - 7 = y , then y x 2 + 2 x y - 7 y = x 2 + 34 x - 71 ⇒ x 2 ( y - 1 ) + x ( 2 y - 34 ) + ( 71 - 7 y ) = 0 If x be real for all values, then 4 ( y - 17 ) 2 - 4 ( y - 1 ) ( 71 - 7 y ) ≥ 0 ⇒ y 2 - 34 y + 289 + 7 y 2 - 78 y + 71 ≥ 0 ⇒ 8 y 2 - 112 y + 360 ≥ 0 or y 2 - 14 y + 45 ≥ 0 ⇒ ( y - 5 ) ( y - 9 ) ≥ 0 i.e. y ≤ 5 ∪ y ≥ 9 Hence, the given expression can not have values between 5 and 9 .

If x be real prove that x 2 + 34 x - 71 x 2 + 2 x - 7 can have no value between 5 and 9 .

Let x 2 + 34 x - 71 x 2 + 2 x - 7 = y , then y x 2 + 2 x y - 7 y = x 2 + 34 x - 71 ⇒ x 2 ( y - 1 ) + x ( 2 y - 34 ) + ( 71 - 7 y ) = 0 If x be real for all values, then 4 ( y - 17 ) 2 - 4 ( y - 1 ) ( 71 - 7 y ) ≥ 0 ⇒ y 2 - 34 y + 289 + 7 y 2 - 78 y + 71 ≥ 0 ⇒ 8 y 2 - 112 y + 360 ≥ 0 or y 2 - 14 y + 45 ≥ 0 ⇒ ( y - 5 ) ( y - 9 ) ≥ 0 i.e. y ≤ 5 ∪ y ≥ 9 Hence, the given expression can not have values between 5 and 9 .

If x be real, the expression ( x + m ) 2 - 4 m n 2 ( x - n ) admits all values except those between 2 n and 2 m .

Let ( x + m ) 2 - 4 m n 2 ( x - n ) = y , then x 2 + 2 m x + m 2 - 4 m n = 2 y x - 2 y n ⇒ x 2 + ( 2 m - 2 y ) x + m 2 - 4 m n + 2 y n = 0 ∵ x is real for all values ∴ 4 ( m - y ) 2 - 4 m 2 - 4 m n + 2 y n should be + ve ⇒ y 2 - 2 y ( m + n ) + 4 m n ≥ 0 ⇒ y ( y - 2 m ) - 2 n ( y - 2 m ) ≥ 0 ⇒ ( y - 2 n ) ( y - 2 m ) ≥ 0 It is + ve for all values of y except those lying between 2 n and 2 m .

If P ( x ) = a x 2 + b x + c and Q ( x ) = - a x 2 + d x + c and a c ≠ 0 . Then prove that the product P ( x ) Q ( x ) = 0 has at least two real roots.

Given that a c ≠ 0 , So let us consider two cases. Case 1 . Let a c > 0 The discriminant of the Quadratic Q ( x ) = - a x 2 + d x + c = 0 is ∆ Q = d 2 + 4 a c Since a c > 0 ⇒ 4 a c > 0 and also d 2 ≥ 0 So ∆ Q = d 2 + 4 a c > 0 , So Q ( x ) = 0 has two distinct real roots and the nature of roots of P ( x ) = 0 can be of any nature. Case 2 . Let a c < 0 The discriminant of the Quadratic P ( x ) = a x 2 + b x + c = 0 is ∆ P = b 2 - 4 a c Since a c < 0 ⇒ - 4 a c > 0 and b 2 ≥ 0 So ∆ P = b 2 - 4 a c > 0 , So P ( x ) = 0 has two distinct real roots and the nature of roots of Q ( x ) can be of any nature. So from both the cases P ( x ) Q ( x ) = 0 has at-least two real roots. Hence the proved.

For what value of m , will the equation x 2 - b x a x - c = m - 1 m + 1 have roots equal in magnitude but opposite in sign?

Given equation is x 2 - b x a x - c = m - 1 m + 1 ⇒ x 2 - b x m + 1 = a x - c m - 1 ⇒ x 2 m + 1 - x b m + b = x a m - a - c m + c ⇒ x 2 m + 1 - b m + b + a m - a x + c m - c = 0 If the roots are equal in magnitude but opposite in sign, then their sum must be zero. (i.e., α + β = 0 ) ∴ b m + b + a m - a m + 1 = 0 ( ∵ Sum of roots α + β = - b a ) ⇒ b m + b + a m - a = 0 ⇒ m b + a = a - b ⇒ m = a - b a + b