[tex]if \: \: \alpha \: and \: \beta \: are \: the \: zeroes \: \: of \\ polynomial \: f(x) = {6x}^{2} + x - 2 \\ find \: the \: value \: of \: \frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta [/tex]
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[tex]if \: \: \alpha \: and \: \beta \: are \: the \: zeroes \: \: of \\ polynomial \: f(x) = {6x}^{2} + x - 2 \\ find \: the \: value \: of \: \frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta [/tex]
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Answer: The required answer is [tex]5/6[/tex].
Explanation:
Given quadratic polynomial is [tex]6x^{2} +x-2[/tex].
Comparing the equation with the general quadratic polynomial [tex]ax^{2} +bx+c[/tex].
Here, we get [tex]a=6,b=1,c=-2.[/tex]
Let [tex]\alpha ,\beta[/tex] be the roots of the given polynomial, then we have
Sum of roots= [tex]\alpha +\beta[/tex] = [tex]-b/a=-1/6.[/tex]
Product of roots= [tex]\alpha \beta[/tex] = [tex]-2/6=-1/3[/tex].
According to question, [tex]1/\alpha +1/\beta -\alpha \beta =(\alpha +\beta )/\alpha \beta -\alpha \beta[/tex]
⇒ [tex](\alpha +\beta )/\alpha \beta =-1*3/-1*6=1/2[/tex]
⇒[tex](\alpha +\beta )/\alpha \beta -\alpha \beta=1/2-(-1/3)=1/2+1/3[/tex]
⇒[tex]1/\alpha +1/\beta -\alpha \beta =3+2/6=5/6[/tex].
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \alpha \: and \: \beta \: are \: zeroes \: of \: polynomial \: f(x) = {6x}^{2} + x - 2 \\ \\ [/tex]
We know,
[tex]\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
[tex]\bf\implies \: \alpha + \beta = - \dfrac{1}{6} \\ \\ [/tex]
Also,
[tex]\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
[tex]\bf\implies \: \alpha\beta = \dfrac{ - 2}{6} = - \frac{1}{3} \\ \\ [/tex]
Now, Consider
[tex]\sf \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } - \alpha \beta \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ \beta + \alpha }{ \alpha \beta } - \alpha \beta \\ \\ [/tex]
[tex]\sf \: = \: \bigg( - \dfrac{1}{6} \bigg) \div \bigg( - \dfrac{1}{3} \bigg) + \dfrac{1}{3} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{6} \times 3 + \dfrac{1}{3} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} + \dfrac{1}{3} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 + 2}{6} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{5}{6} \\ \\ [/tex]
Hence,
[tex]\sf \:\bf\implies \:\frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta = \: \dfrac{5}{6} \\ \\ [/tex]