Question

If ad  bc, then prove that the equation (a2 + b2 ) x2 + 2 (ac + bd) x + (c2 + d2 ) = 0 has no real roots.

Answer 1

I hope you will understand

Answer 2

Quadratic equation:

$( {a}^{2} + {b}^{2} ) {x}^{2} + 2(ac + bd)x + ( {c}^{2} + {d}^{2} ) = 0$

Here,

$a = ( {a}^{2} + {b}^{2} ) \\ = {a}^{2} + {b}^{2}$

$b = 2(ac + bd) \\ = 2ac + 2bd$

$c = ( {c}^{2} + {d}^{2} ) \\ = {c}^{2} + {d}^{2}$

Given that,

The quadratic equation has no real roots,

Therefore, $D < 0$

Now we know that,

$D = {b}^{2} - 4ac$

Therefore,

$b {}^{2} - 4ac < 0$

$(2ac + 2bd) {}^{2} - 4(a {}^{2} + {b}^{2} )( {c}^{2} + {d}^{2} ) < 0$

$(4 {a}^{2} {c}^{2} + 4 {b}^{2} {d}^{2} + 8abcd) + ( - 4 {a}^{2} {c}^{2} - 4 {a}^{2} {d}^{2} - 4 {b}^{2} {c}^{2} - 4 {b}^{2} {d}^{2} ) < 0$

$4 {a}^{2} {c}^{2} + 4 {b}^{2} {d}^{2} + 8abcd - 4 {a}^{2} {c}^{2} - 4 {a}^{2} {d}^{2} - 4 {b}^{2} {c}^{2} - 4 {b}^{2} {d}^{2} < 0$

$(4 {a}^{2} {c}^{2} - 4 {a}^{2} {c}^{2} ) +( 4 {b}^{2} {d}^{2} - 4 {b}^{2} {d}^{2}) + 8abcd - 4 {a}^{2} {d}^{2} - 4 {b}^{2} {c}^{2} < 0$

$8abcd - 4 {a}^{2} {d}^{2} - 4 {b}^{2} {c}^{2} < 0$

$- 4 {a}^{2} {d}^{2} - 4 {b}^{2} {c}^{2} + 8abcd < 0$

$- 4( {a}^{2} {d}^{2} + {b}^{2} {c}^{2} - 2abcd) < 0$

${a}^{2} {d}^{2} + {b}^{2} {c}^{2} - 2abcd < \frac{0}{ - 4}$

${a}^{2} {d}^{2} + {b}^{2} {c}^{2} - 2abcd < 0$

$(ad - bc) {}^{2} < 0$

$ad - bc < \sqrt{0}$

$ad - bc <0$

If, $ad = bc$then, $D= 0$

Since, $D < 0$ ( equation has no real roots)

$ad ≠bc$