Question

If the equation (a² + b²)x²- 2b(a + c)x+b² + c² = 0 has equal roots then (a) 2b =a +c (6) b² = ac (c) b= 2ac/a+c (d) b=ac
Step By Step Answer required.​​

Answer 1

Step-by-step explanation:

For a quadratic equation,

$a {x}^{2} + bx \: + c = 0$

If roots are equal, then

$discriminant \: \: {b}^{2} - 4ac \: = 0$

Therefore,

${ (- 2b(a \: + c))}^{2} - 4( {a}^{2} + {b}^{2} )( {b}^{2} + {c}^{2} ) = 0$

=

$4 {b}^{2} ( {a \: + c)}^{2} - 4( {a}^{2} {b}^{2} + {a}^{2} {c}^{2} + {b}^{4} + {b}^{2} {c}^{2} ) = 0$

=

${b}^{2} ( {a}^{2} + {c}^{2} + 2ac) - {a}^{2} {b}^{2} - {a}^{2} {c}^{2} - {b}^{4} - {b}^{2} {c}^{2} = 0$

=

${b}^{2} {a}^{2} + {b}^{2} {c}^{2} + 2ac {b}^{2} - {a}^{2} {b}^{2} - {a}^{2} {c}^{2} - {b}^{4} - {b}^{2} {c}^{2} = 0$

=

$2ac {b}^{2} - {a}^{2} {c}^{2} - {b}^{4} = 0$

=

${(ac - {b}^{2}) }^{2} = 0$

=

${b}^{2} = ac$

Hence, option (b) is correct.