for what valu of k, will 7/3 be a root of
3x square -13x-k=0?
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f(x)=3x²-13x-k
f(7/3)=3(7/3)²-13(7/3)-k=0
49/3-21/3-k=0
28/3=k
f(7/3)=3(7/3)²-13(7/3)-k=0
49/3-21/3-k=0
28/3=k
Answered by
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K is the root so x=7/3
3x^2-13x-k=0
3 (7/3)^2-13 (7/3)-k=0
49/3-91/3-k=0
-42/3-k=0
k=-42/3
3x^2-13x-k=0
3 (7/3)^2-13 (7/3)-k=0
49/3-91/3-k=0
-42/3-k=0
k=-42/3
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