Math, asked by mridulmohnsingh, 1 month ago

obtain all the zeros of 3x³-x²-3x+1.
given that one of its zeros is 1​

Answers

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:Let \: remaining \: zeroes \: be \: a \: and \: b

So,

We have,

\rm :\longmapsto\:1,a,b \: are \: zeroes \: of \:  3{x}^{3} -  {x}^{2} - 3x + 1

We know,

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}

\bf\implies \:1 + a + b =  - \dfrac{( - 1)}{3}  = \dfrac{1}{3}

\bf\implies \:a + b =  - \dfrac{2}{3}  -  -  - (1)

Also,

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{3}}}}

\rm :\longmapsto\:1 \times a \times b =  - \dfrac{1}{3}

\rm :\longmapsto\:ab = -  \dfrac{1}{3}

\rm :\longmapsto\:3ab = -  1

\rm :\longmapsto\:3a\bigg( - \dfrac{2}{3} - a \bigg)  =  - 1 \:  \: \:  \:  \:  \:   \{using \: (1) \}

\rm :\longmapsto\: - 2a - 3 {a}^{2}  =  - 1

\rm :\longmapsto\: {3a}^{2} + 2a - 1 = 0

\rm :\longmapsto\: {3a}^{2} + 3a - a - 1 = 0

\rm :\longmapsto\:3a(a + 1) - 1(a + 1) = 0

\rm :\longmapsto\:(3a  -  1)(a  +  1) = 0

\bf\implies \:a =  - 1 \:  \:  \: or \:  \:  \: a =   \dfrac{1}{3}

Put the values of a in equation (1) to get values of b.

\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf b =  - \dfrac{2}{3} - a  \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf  - 1 & \sf  \dfrac{1}{3}  \\ \\ \sf   \dfrac{1}{3}  & \sf  - 1  \end{array}} \\ \end{gathered}

Hence,

\bf :\longmapsto\:Zeroes \: are \: 1, - 1, \dfrac{1}{3}

Additional Information :-

\rm :\longmapsto\:If \:  \alpha  \: and \:  \beta  \: are \: zeroes \: of \:  {ax}^{2} + bx + c \: then

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

and

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

Answered by muskanshi536
2

Step-by-step explanation:

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:Let \: remaining \: zeroes \: be \: a \: and \: b

So,

We have,

\rm :\longmapsto\:1,a,b \: are \: zeroes \: of \:  3{x}^{3} -  {x}^{2} - 3x + 1

We know,

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}

\bf\implies \:1 + a + b =  - \dfrac{( - 1)}{3}  = \dfrac{1}{3}

\bf\implies \:a + b =  - \dfrac{2}{3}  -  -  - (1)

Also,

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{3}}}}

\rm :\longmapsto\:1 \times a \times b =  - \dfrac{1}{3}

\rm :\longmapsto\:ab = -  \dfrac{1}{3}

\rm :\longmapsto\:3ab = -  1

\rm :\longmapsto\:3a\bigg( - \dfrac{2}{3} - a \bigg)  =  - 1 \:  \: \:  \:  \:  \:   \{using \: (1) \}

\rm :\longmapsto\: - 2a - 3 {a}^{2}  =  - 1

\rm :\longmapsto\: {3a}^{2} + 2a - 1 = 0

\rm :\longmapsto\: {3a}^{2} + 3a - a - 1 = 0

\rm :\longmapsto\:3a(a + 1) - 1(a + 1) = 0

\rm :\longmapsto\:(3a  -  1)(a  +  1) = 0

\bf\implies \:a =  - 1 \:  \:  \: or \:  \:  \: a =   \dfrac{1}{3}

Put the values of a in equation (1) to get values of b.

\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf b =  - \dfrac{2}{3} - a  \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf  - 1 & \sf  \dfrac{1}{3}  \\ \\ \sf   \dfrac{1}{3}  & \sf  - 1  \end{array}} \\ \end{gathered}

Hence,

\bf :\longmapsto\:Zeroes \: are \: 1, - 1, \dfrac{1}{3}

Additional Information :-

\rm :\longmapsto\:If \:  \alpha  \: and \:  \beta  \: are \: zeroes \: of \:  {ax}^{2} + bx + c \: then

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

and

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

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