Question

If the roots of the equation (a² +b² ) x² + 2(bc-ad)x+c²+d²=0 are equal, show that ac +bd=0.

Answer 1

Answer:

For equal root discriminate of above equation must be zero i.e.

$for \: \: p {x}^{2} + qx + r = 0 \\ {q}^{2} - 4pr = 0 \\ \\ (2(bc - ad)) {}^{2} - 4( {a}^{2} + {b}^{2} )( {c}^{2} + {d}^{2} ) = 0 \\ \\ 4 {b}^{2} {c}^{2} + 4 {a}^{2} {d}^{2} - 8abcd - 4 {a}^{2} {c}^{2} - 4 {d}^{2} {a}^{2} \\ - 4 {b}^{2} {c}^{2} - 4 {b}^{2} {d}^{2} = 0 \\ \\ - 4 {a}^{2} {c}^{2} - 4 {b}^{2} {d}^{2} - 8abcd = 0 \\ \\ divide \: \: by \: \: - 4\\ (ac) {}^{2} + (bd) ^2+ 2(ac)(bd) = 0 \\ \\ (ac + bd) { }^{2} = 0 \\ \\ \pink{ac + bd = 0}$