if x+3 one of the root of equation x square - 2kx-6=0 then find valur of k
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The root of quadratic equation is x+3 ....
Equation is x²-2kx-6=0
Now,substitute x+3 in eq.
x²-2kx-6=0....
therefore , (x+3)²-2k(x+3)-6=0
x²+6x+9-2kx-6k-6=0
x²+6x-2kx-6k+3=0
x²+6x-6k+3=2kx
2k=(x²+6x-6k+3)÷x
2k+6k=x+6x+3
8k=7x+3
k=(7x+3)÷8
we know that , x=x +3
therefore , k=(7(x+3)+3)÷8
k = (7x+21+3)÷8
k= (7x+24)÷8
therefore, The value of k is 7x+24/8
Equation is x²-2kx-6=0
Now,substitute x+3 in eq.
x²-2kx-6=0....
therefore , (x+3)²-2k(x+3)-6=0
x²+6x+9-2kx-6k-6=0
x²+6x-2kx-6k+3=0
x²+6x-6k+3=2kx
2k=(x²+6x-6k+3)÷x
2k+6k=x+6x+3
8k=7x+3
k=(7x+3)÷8
we know that , x=x +3
therefore , k=(7(x+3)+3)÷8
k = (7x+21+3)÷8
k= (7x+24)÷8
therefore, The value of k is 7x+24/8
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