Question

show that if root of the following quadratic equation are equation are equal, then ad=bc. x2(a2+b2)+2(ac+bd)+(c2+d2)=0

Answer 1

$( {a}^{2} + {b}^{2} ) {x}^{2} + 2(ac + bd)x + ( {c}^{2} + {d}^{2} ) = 0$
has equal roots. This means that the discriminant equals zero.

Thus,

$(2(ac + bd))^{2} - 4( {a}^{2} + {b}^{2} )( {c}^{2} + {d}^{2} ) = 0 \\ 4(ac) ^{2} + 8abcd + 4 {(bd)}^{2} - 4( ac)^{2} - 4(ad)^{2} - 4(bc) ^{2} - 4 (bd)^{2} = 0 \\ - 4(ad) ^{2} - 4(bc) ^{2} + 8abcd = 0 \\ (ad) ^{2} + (bc) ^{2} - 2(ad)(bc) = 0 \\ (ad - bc) ^{2} = 0 \\ ad-bc=0 \\ \boxed{ad = bc}$

Answer 2

2 roots are equal only if discriminant, b²-4ac=0
By question if roots are equal the ad=bc therefore we can say that the two roots are real and equal.
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