Question

if roots of the eq. (a2+b2)×X2-2b(a+c)x+(b2+c2)=0 are equal then find the value of b2

Answer 1

SOLUTION ☺️

$= > ( {a}^{2} + {b}^{2} ) {x}^{2} - 2b(a + c)x + ( {b}^{2} + {c}^{2} ) = 0 \\ = > on \: comparing \: with \: a \: general \: quadratic \: equation \\ = > ax {}^{2} + b {x} + c = 0 \: we \: get \\ = > a = ( {a}^{2} + {b}^{2} ) \\ = > b = - 2b(a + c) \\ = > c = ( {b}^{2} + {c}^{2} ) \\ \\ = > discriminate = {b}^{2} - 4ac \\ as \: the \: roots \: are \: real \: and \: equal \: \\ = > therefore \: d = 0 \\ = > ( - 2b(a + c)) {}^{2} = 4( {a}^{2} + {b}^{2} )( {b}^{2} + {c}^{2} ) \\ = > {b}^{2} ( {a}^{2} + 2ac + {c}^{2} ) = {a}^{2} {b}^{2} + {a}^{2} {c}^{2} + {b}^{4} + {b}^{2} {c}^{2} \\ = > (ac) {}^{2} - 2(ac)( {b}^{2} ) + ( {b}^{2} ) {}^{2} = 0 \\ = > (ac - {b}^{2} ) {}^{2} = 0 \\ = > (ac - {b}^{2} ) = 0 \\ = > ac = {b}^{2}$

HOPE it helps ✔️

Answer 2

see the attachment.....