Question

QUADRATIC EQUATIONS
4.43
quation (a? + b²) x2 - 2(ac + bd)x +(c^2 + d^2) = 0 are equal, prove that
a/b=c/d​

Answer 1

We have

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

Compare with

$ax {}^{2} + bx + c = 0$

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

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

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

if roots are equal

$d = \: b {}^{2} - 4ac \: or \: b {}^{2} = 4ac$

$- 2(ac + bd ) {}^{2} \: = \:( a { }^{2} + b {}^{2}) + \\( c {}^{2} + d {}^{2} )$

$a {}^{2} c { }^{2} + 2acd + b {}^{2} d {}^{2} = 4(a {}^{2} c {}^{2} \\ + \: a {}^{2} d {}^{2} \: b {}^{2} c {}^{2} \: b {}^{2} d {}^{2} )$

$2acd \: = a {}^{2} d {}^{2} + b {}^{2} c {}^{2}$

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

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

$0 = ad = bc$

$ad = bc$

$a \div b = c \div d$

$hence \: proved$

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