Question

Solve 26th question.

Answer 1

Here is your answer :

Given,

Quadratic equation = ( 2k + 1 )x² - ( 7k + 2 )x + ( 7k - 3 ) = 0

Here,

Coefficient of x² ( a ) = ( 2k + 1 )

Coefficient of x ( b ) = - ( 7k + 2 )

Constant term ( c ) = ( 7k - 3 )

For any quadratic equation to have equal roots, its Discriminat ( D = b² - 4ac ) should be equal to zero.

=> D = 0

=> b² - 4ac = 0

Substitute the value of a, b and c.

=> [ -( 7k + 2 ) ]² - 4( 2k + 1 )( 7k - 3) = 0

=> ( -7k )² + ( -2 )² + 2( -7k ) ( -2 ) - ( 8k + 4 ) ( 7k - 3 ) = 0

=> 49k² + 4 + 28k -8k( 7k - 3 ) - 4( 7k - 3 ) = 0

=> 49k² + 4 + 28k - 56k² + 24k - 28k + 12 = 0

=> 49k² - 56k² + 28k + 24k - 28k + 4 + 12 = 0

=> -7k² + 24k + 16 = 0

=> - ( 7k² - 24k - 16 ) = 0

=> ( 7k² - 24k - 16 ) = 0

=> 7k² - 28k + 4k - 16 = 0

=> 7k( k - 4 ) + 4( k - 4 ) = 0

=> ( k - 4 ) ( 7k + 4 ) = 0

=>(k-4)= 0/(7k + 4 ) , ( 7k + 4 ) = 0/( k - 4 )

=> ( k - 4 ) = 0 , ( 7k + 4 ) = 0

=> k = 4 , 7k = -4

•°• k = 4 , ( -4/7 )

Now,

Quadratic equation = ( 2k + 1)x² - ( 7k + 2 )x + ( 7k - 3 ) = 0

Substitute k = 4,

=> ( 2 × 4 + 1 )x² - ( 7 × 4 + 2 )x + ( 7 × 4 - 3 ) = 0

=> ( 8 + 1 )x² - ( 28 + 2 )x + ( 28 - 3 ) = 0

=> 9x² - 30x + 25 = 0

=> ( 3x )² - 2 × ( 3x ) ( 5 ) + ( 5 )² = 0

Using identity :

[ a² - 2ab + b² = ( a - b )² ]

=> ( 3x - 5 )² = 0

=> ( 3x - 5 ) = √0

=> ( 3x - 5 ) = 0

=> 3x = 5

•°• x = ( 5/3 )

Zeroes = ( 5/3 ) and ( 5/3 ).

When, k = ( -4/7 )

=> ( 2k + 1 )x² - ( 7k + 2 )x + ( 7k - 3 ) = 0

Substitute k = ( -4/7 ),

=> [ 2( -4/7 ) + 1 ]x² - [ 7( -4/7 ) + 2 ]x + [ 7( -4/7 ) - 3 ] = 0

=> [ ( -8/7 ) + 1 ]x² - [ -4 + 2 ]x + ( -4 - 3 ) = 0

=> ( -1/7 )x² - ( -2 )x + ( -7 ) = 0

=> ( -1/7 )x² + 2x - 7 = 0

=> - [ ( 1/7 )x² -2x + 7 ] = 0

=> ( 1/7 )x² - 2x + 7 = 0

=> [ ( 1/√7 )x ]² - 2 [ (1/√7 )x ] ( √7 ) + ( √7 )² = 0

Using identity :

=> [ a² - 2ab + b² = ( a - b )² ]

=> [ ( 1/√7 )x - √7 ]² = 0

=> [ ( 1/√7 )x - √7 ] = 0

=> ( 1/√7 )x = √7

=> x = √7 × √7

•°• x = 7

Zeroes = 7 , 7

If k = 4 then zeroes are (5/3 ) and ( 5/3 ).

If , k = ( -4/7 ) then zeroes are ( 7 ) and ( 7 ).




Hope it helps !!

Answer 2

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HERE IS YOUR ANSWER ☞
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$given \: quadratic \: equation \: = > \\ \\ = > (2k + 1) {x}^{2} - (7k + 2)x + (7k - 3) = 0 \\ \\ here \\ = > a \: = (2k + 1) \\ = > b = (7k + 2) \\ = > c = (7k - 3) \\ \\ now \\ \\ = > d = {b}^{2} - 4ac \\ \\ = > d = {(7k + 2)}^{2} - 4 \times (2k + 1)(7k - 3) \\ \\ = > d = 49 {k}^{2} + 4 + 28 - 4(14 {k}^{2} + 7k - 6k - 3) \\ \\ = > d = - 7 {k}^{2} + 24k + 16 \\ \\ we \: know \: that \: for \: equal \: roots \: \\ \\ = > d = 0 \\ \\ = > - 7 {k}^{2} + 24k + 16 = 0 \\ \\ = > 7 {k}^{2} - 28k \: + 4k - 16 = 0 \\ \\ = > 7k(k - 4) + 4(k - 4) = 0 \\ \\ = > (7k + 4)(k - 4) = 0 \\ \\ then \\ \\ = > k = - \frac{4}{7} \: \: \: or \: \: \: \: k = 4 \\ \\ now \\ \\ plug \: the \: value \: of \: of \: = > \: k = 4 \: in \: equation \: we \: get \\ \\ = > (2k + 1) {x}^{2} - (7k + 2)x + (7k - 3) = 0 \\ \\ = > 9 {x}^{2} - 30x + 25 = 0 \\ \\ = > {(3x - 5)}^{2} = 0 \: \: \: \: \: \: ( {a}^{2} - 2ab + {b = {(a - b)}^{2} }) \\ \\ = > 3x - 5 = 0 \\ \\ = > x = \frac{5}{3} \\ \\ = > zeros = \frac{5}{3} \: and \: \frac{5}{3} \\ \\ now \\ \\ plug \: value \: = > k = - \frac{4}{7} \\ \\ = > (2( - \frac{4}{7} + 1) {x}^{2} - (7( - \frac{4}{7} ) + 2)x + (7( - \frac{4}{3} ) - 3) = 0 \\ \\ = > ( - \frac{1}{7} ) {x}^{2} + 2x - 7 = 0 \\ \\ = > \frac{1}{7} {x}^{2} - 2x + 7 = 0 \\ \\ = > {( \frac{1}{ \sqrt{7} }) }^{2} - 2( \frac{1}{ \sqrt{7} } x) \sqrt{7} + { \sqrt{7} }^{2} = 0 \\ \\ = > {( \frac{1}{ \sqrt{7} }x - \sqrt{7} )}^{2} = 0 \\ \\ = > \frac{1}{ \sqrt{7} } x = \sqrt{7 } \\ \\ = > x = 7 \\ \\ = > zeros \: = 7 \: and \: 7$


if "k=4" then roots are "5/3 and 5/3"

if "k=-4/7" then roots are 7 and 7
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