5.If a and ß are the zeroes of 6x² + x-2 . Find the value of a/B+ B/

1.	5.If a and ß are the zeroes of 6x² + x-2 . Find the value of a/B+ B/a.
Answer» Answer The VALUE of α/β + β/α is -25/12 $\bf \large \underline{Given : }$ The QUADRATIC POLYNOMIAL 6x² + X - 2 α and β are two zeroes of the given polynomial $\bf \large \underline{To \: Find : }$ The value of α/β + β/α $\bf \large \underline{Solution : }$ We have the polynomial , 6x² + x - 2 and zeroes are α and β Therefore , from the sum relation of zeroes $\sf sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: {x}}{coefficient \: of \: {x}^{2} } \\ \\ \implies \sf\alpha + \beta = - \dfrac{1}{6} \\ \\ \sf \implies {( \alpha + \beta )}^{2} = \frac{1}{36} \\ \\ \sf \implies { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{1}{36} \longrightarrow(1)$ And from the relation product of the zeroes and coefficients $\sf product \: of \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$ $\sf\implies \alpha\beta = \dfrac{-2}{6} \\\\ \sf\implies \alpha \beta = -\dfrac{1}{3} \longrightarrow (2)$ DIVIDING (2) by (1) we have , $\sf\implies \dfrac{\alpha^{2}}{\alpha\beta}+\dfrac{\beta^{2}}{\alpha\beta}+ \dfrac{2\alpha\beta}{\alpha\beta} = \dfrac{\dfrac{1}{36}}{-\dfrac{1}{3}} \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + 2 = -\dfrac{1}{12} \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = -\dfrac{1}{12} - 2 \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{-1 - 24}{12} \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = -\dfrac{25}{12}$ Thus , the required value of α/β + β/α is -25/12

5.If a and ß are the zeroes of 6x² + x-2 . Find the value of a/B+ B/a.

Answer»

Answer

The VALUE of α/β + β/α is -25/12

$\bf \large \underline{Given : }$

The QUADRATIC POLYNOMIAL

6x² + X - 2
α and β are two zeroes of the given polynomial

$\bf \large \underline{To \: Find : }$

The value of

α/β + β/α

$\bf \large \underline{Solution : }$

We have the polynomial ,

6x² + x - 2

and zeroes are α and β

Therefore , from the sum relation of zeroes

$\sf sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: {x}}{coefficient \: of \: {x}^{2} } \\ \\ \implies \sf\alpha + \beta = - \dfrac{1}{6} \\ \\ \sf \implies {( \alpha + \beta )}^{2} = \frac{1}{36} \\ \\ \sf \implies { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{1}{36} \longrightarrow(1)$

And from the relation product of the zeroes and coefficients

$\sf product \: of \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\sf\implies \alpha\beta = \dfrac{-2}{6} \\\\ \sf\implies \alpha \beta = -\dfrac{1}{3} \longrightarrow (2)$

DIVIDING (2) by (1) we have ,

$\sf\implies \dfrac{\alpha^{2}}{\alpha\beta}+\dfrac{\beta^{2}}{\alpha\beta}+ \dfrac{2\alpha\beta}{\alpha\beta} = \dfrac{\dfrac{1}{36}}{-\dfrac{1}{3}} \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + 2 = -\dfrac{1}{12} \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = -\dfrac{1}{12} - 2 \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{-1 - 24}{12} \\\\ \sf\implies \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = -\dfrac{25}{12}$

Thus , the required value of α/β + β/α is -25/12