Question

One root of the quadratic equation kx2-10x+3=0 is 1/3

Answer 1

Answer:

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Step-by-step explanation:

since one of the zero of the polynomial is 1/3

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0k - 30 +27 =0

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0k - 30 +27 =0k -3 =O

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0k - 30 +27 =0k -3 =O=> k =3

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0k - 30 +27 =0k -3 =O=> k =3Step-by-step explanation:

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0k - 30 +27 =0k -3 =O=> k =3Step-by-step explanation:hope it helps you

since one of the zero of the polynomial is 1/3substitute 1/3 in place of "x" in the polynomial .so,,,kx2 -10 x +3 =0k (1/3)^2 -10(1/3) +3 =0k/9 -10/3 +3 =0k/9 -30/9 + 27/9 =0k - 30 +27 / 9 =0k - 30 +27 =0k -3 =O=> k =3Step-by-step explanation:hope it helps youmark as brainliest