Question

27. if a and B are the zeros of the quadratic polynomial f(x) = 6x2 + x - 2, find the value of α/β+β/α
​

Answer 1

Answer:-

Given Polynomial: 6x² + x - 2

Let a = 6 ; b = 1 ; c = - 2

We know that,

Sum of the zeroes = - b/a

→ α + β = - 1/6 -- equation (1)

Product of the zeroes = c/a

→ αβ = - 2/6

→ αβ = - 1/3 -- equation (2)

We have to find:

α/β+β/α = ?

Taking LCM we get,

→ ( α² + β² ) / αβ

We know that,

(a + b)² = a² + b² + 2ab

→ (a + b)² - 2ab = a² + b²

→ (α + β)² - 2αβ = α² + β²

Hence,

→ [ (α + β)² - 2αβ ] / αβ

Putting the values from equation (1) and (2) we get,

→ [ ( - 1/6)² - 2( - 1/3) ] / (- 1/3)

→ [ 1/36 + 2/3 ] / ( - 1/3)

→ [ (1 + 24) / 36 ] * ( - 3)

→ 25 / 36 * ( - 3)

→ - 25 / 12

Hence, the value of α/β + β/α is - 25/12.

Answer 2

$\large\mathcal{\underbrace{QUESTION:-}}$

✍️ If $\rm{\alpha}$ and $\rm{\beta}$ are the zeros of the quadratic polynomial f(x) = 6x² + x - 2, find the value of ‘$\rm{\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}}$’ .

$\huge\mathcal{\underbrace{SOLUTION:-}}$

GIVEN :-

• $\rm{\red{\alpha\:and\:\beta}}$ are the zeros of the quadratic polynomial f(x) .

Where,

• f(x) = 6x² + x - 2

CALCULATION :-

✍️ Here,

• coefficient of x² = 6

• coefficient of x = 1

• constant term = -2

$\checkmark\:\mathcal{\purple{Sum\:of\:zeros\:=\:\dfrac{-\:Coefficient\:of\:x\:}{Coefficient\:of\:x^2}\:}}$

$\rm{\implies\:\alpha\:+\:\beta\:=\:\dfrac{-1}{6}\:}$

$\checkmark\:\mathcal{\purple{Product\:of\:zeros\:=\:\dfrac{Constant\:term}{Coefficient\:of\:x^2}\:}}$

$\rm{\implies\:\alpha\:\beta\:=\:\dfrac{-2}{6}\:}$

$\rm{\implies\:\alpha\:\beta\:=\:\dfrac{-1}{3}\:}$

✍️ Now,

$\rm{\:\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:}$

$\rm{=\:\dfrac{\alpha^{2}\:+\:\beta^{2}}{\alpha\:\beta}\:}$

$\rm{=\:\dfrac{(\alpha\:+\:\beta)^2\:-\:2\alpha\beta}{\alpha\:\beta}\:}$

$\rm{=\:\dfrac{(\dfrac{-1}{6})^2\:-\:2\times{\dfrac{-1}{3}}\:}{\dfrac{-1}{3}}\:}$

$\rm{=\:\dfrac{\dfrac{1}{36}\:+\:\dfrac{2}{3}\:}{\dfrac{-1}{3}}\:}$

$\rm{=\:\dfrac{\dfrac{1\:+\:24}{36}}{\dfrac{-1}{3}}\:}$

$\rm{=\:\dfrac{\dfrac{25}{36}}{\dfrac{-1}{3}}\:}$

$\rm{=\:\dfrac{25}{36}\:\times{-3}\:}$

$\rm{=\:\dfrac{-25}{12}\:}$

$\bigstar\:\rm{\green{\boxed{\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:=\:\dfrac{-25}{12}\:}}}$