Question

If $\alpha and \beta$ are the zeros of the quadratic polynomial $f(x)= 6^{2}+x-2$,find the value of $\frac{\alpha}{\beta} +\frac{\beta}{\alpha}$

Answer 1

SOLUTION :

Given : α and β are the zeroes of the quadratic polynomial f(x)= 6x² + x - 2.

On comparing with ax² + bx + c,

a = 6  , b= 1  , c= -2

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = -1/6

α + β = −1/6 ……………………………(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = -2/6 = -1/3

αβ =  −1/3 ………………………….(2)

Now,

α/β+β/α =( α² + β²)/αβ

As we know that ,  α + β)² = α² + β² +  2αβ ,  α² + β² = (α + β)² - 2αβ

α/β+β/α = (α + β)²  –2αβ / αβ

By Substituting the value from eq 1 & 2

= (-⅙)² - 2× -⅓ / -⅓

=(1/36 + ⅔ )/ -⅓

=[ (1/36 + 2×12)/36 ] / -⅓

= [(1/36 +24)/36] / -1/3

= [(1+24)/36] / -⅓

= 25/36 / -⅓

= 25/36 × - 3/1  

= -25/12

Hence, the value of α/β+β/α is -25/12

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Given : α and β are the zeroes of the quadratic polynomial f(x)= x² + x - 2

On comparing with ax² + bx + c,

a = 1 , b= 1 , c= -2

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = - 1/1 = -1

α+β = - 1 ……………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = -2/1 = -2

αβ = - 2 ……………………….(2)

1/α  -  1/ β  
= (β–α)/αβ = -(α -  β) /αβ ……………….(3)

As we know that , (α - β)² = (α+β)² - 4αβ

(α - β)² = - 1² - 4 × -2 = 1 +8 = 9

(α - β)² = 9

(α - β) = √9 =  ±3

(α - β) = ±3 ………………(4)  

By Substituting the value from eq 2 & 4 ,in eq 3 , we get  

(β–α)/αβ = -(α -  β) /αβ

= - ( ±3 ) /-2 =  ±3/2  

1/α  -  1/ β   =  ±3/2  

Hence , the value of   1/α  -  1/ β  =  ± 3/2
HOPE THIS ANSWER WILL HELP YOU….


$aman \: king$

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