Question

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

Answer 1

Answer:

6x^2 +x—2 =0

a and b are zeroes of QUADRATIC EQUATION

6x^2 +4x--3x -2 =0 ( 6× -2 =--12 4×--3= --12 +4--3 =1)

2x(3x +2 ) -- 1(3x+2) =0

(2x-1 ) (3x+-2) =0

2x=1. x=1/2

3x=--2

x= --2/3

a = --2/3. b. . =1/2

Find the value of a/b +b/a

a/b = -- 4/3 b/ a = --3/4

a/b+b/a = -- 4/3 -- 3/4 = (—16 --9) /12 = --25 /12

The required value is --25/12

Answer 2

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