find the value of k for which a root of quadratic equation
x2-2(1+3k)x+7 (3+2k)=0 is 7.
Answers
Answered by
15
Hi ,
x² - 2 ( 1 + 3k ) x + 7( 3 + 2k ) = 0 is a
given quadratic equation whose one
root is 7
Put x = 7 in the equation.
7² - 2 ( 1 + 3k ) 7 + 7 ( 3 + 2k ) = 0
Divide each term with 7 both sides of
the equation ,
7 - 2 ( 1 + 3k ) + 3 + 2k = 0
7 - 2 - 6k + 3 + 2k = 0
- 4k + 8 = 0
-4k = - 8
k = ( - 8 ) / ( - 4 )
k = 2
I hope this helps you.
:)
x² - 2 ( 1 + 3k ) x + 7( 3 + 2k ) = 0 is a
given quadratic equation whose one
root is 7
Put x = 7 in the equation.
7² - 2 ( 1 + 3k ) 7 + 7 ( 3 + 2k ) = 0
Divide each term with 7 both sides of
the equation ,
7 - 2 ( 1 + 3k ) + 3 + 2k = 0
7 - 2 - 6k + 3 + 2k = 0
- 4k + 8 = 0
-4k = - 8
k = ( - 8 ) / ( - 4 )
k = 2
I hope this helps you.
:)
Answered by
7
The root for this equation is 7 which means x=7
so if x =7,then
the eq.= (7)^2 -2(1+3k)7+7(3+2k)=0
..............= 49-14-42k+21+14k=0
.............= 70-14-28k=O
.............=28k=56
.............=k=56/28
.............k=2.
so if x =7,then
the eq.= (7)^2 -2(1+3k)7+7(3+2k)=0
..............= 49-14-42k+21+14k=0
.............= 70-14-28k=O
.............=28k=56
.............=k=56/28
.............k=2.
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