if one root of the quadratic epquation x²-7x+k=0is4
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to find k,
x²-7x+k=0
since 4 is a zero, substituting:
(4)²-7(4)+k=0
16-28+k=0
-12+k=0
k=12
x²-7x+k=0
since 4 is a zero, substituting:
(4)²-7(4)+k=0
16-28+k=0
-12+k=0
k=12
Answered by
1
x²- 7x + k = 0 x= 4 given
16-28 + K =0
k = 12
x²-7x+ 12
x² -4x - 3x +12
x(x-4)-3(x-4)
(x-3)(x-4) = 0
x = 3 ,x = 4
mark as brainiest please
16-28 + K =0
k = 12
x²-7x+ 12
x² -4x - 3x +12
x(x-4)-3(x-4)
(x-3)(x-4) = 0
x = 3 ,x = 4
mark as brainiest please
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