Question

If \[\left( a^2 + b^2 \right) x^2 + 2\left( ab + bd \right)x + c^2 + d^2 = 0\] has no real roots, then

पर्याय

• ab = bc

• ab = cd

• ac = bd

• ad ≠ bc

Answer 1

The given quadric equation is  \[\left( a^2 + b^2 \right) x^2 + 2\left( ab + bd \right)x + c^2 + d^2 = 0\] , and roots are equal.

Here, `a = (a^2 + b^2 ),b = 2 (ab + bd) and , c = c^2 + d^2`

As we know that `D = b^2 - 4ac`

Putting the value of  `a = (a^2 + b^2 ),b = 2 (ab + bd) and , c = c^2 + d^2`

`={2 (ab + bd)}^2 - 4 xx (a^2 _b^2) xx (c^2 + d^2)`

` = 4a^2b^2 + 4b^2d^2 + 8ab^2d - 4(a^2c^2 + a^2 d^2 +b^2c^2 + b^2 d^2)`

`=4a^2b^2 + 4b^2d^2 + 8ab^2d - 4a^2c^2 - 4a^2d^2 - 4b^2 c^2 - 4b^2d^2`

`= 4a^2b^2 + 8ab^2 d - 4a^2c^2 - 4a^2d^2 - 4b^2c^2`

`= 4 (a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2)`

The given equation will have no real roots, if D < 0

`4 (a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2) < 0`

`a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2) < 0`