Question

find the value of c for which quadratic equation 4x square -2 ( C + 1 ) X +(c+ 4 )has equal roots

Answer 1

$\bold {Equation : 4x^{2} - 2( c + 1 ) x + ( c + 4 ) } \\ \\ \\ \mathbf{ On \: \: comparing \: \: the \: \: given \: \: equation \: \: with \: \: ( ax^{2} + bx + c ) , we \: \: get \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [ a, b \: \: and \: \: c \: \: are \: \: variables \: \: which \: \: are \: \: taken \: \: to \: \: show \: \: the \: \: quadratic equation ] \: \: } \\ \\ a = 4 \\ b = -2( c + 1 ) \\ c = ( c + 4 )$





We know, Discriminant = b^{2} - 4ac


=> [ -2( c + 1 ) ]² - 4[ 4 × ( c + 4 )


=> [ 4( c + 1 )² ] - 4( 4c + 16 )


=> 4( c² + 1 + 2c ) - 4( 4c + 16 )


=> 4( c² + 1 + 2c - 4c - 16 )


=> 4( c² + 1 - 16 + 2c - 4c )


=> 4( c² - 15 - 2c )



$\large{ Note }: For \: \: equal \: \: roots, \: \: discriminant = 0$





Hence,

=> 4( c² - 15 - 2c ) = 0


=> c² - 2c - 15 = 0


=> c² - ( 5 - 3 ) c - 15 = 0


=> c² - 5c + 3c - 15 = 0


=> c( c - 5 ) + 3( c - 5 ) = 0


=> ( c - 5 ) ( c + 3 ) = 0



$\text{By Zero Product Rule, }$




=> c = 5 or c = -3