Question

If α and ß are the zeroes of the quadratic polynomial

2

- − 4, find the value of 1



+

1

ß

–

αß.​

Answer 1

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Given that alpha and beta are the zeroes of polynomial x² + x - 6

we've to find the value of 1/alpha + 1/beta

so let's find the zeroes of the given polynomial first.

using splitting the middle term method,

➡ x² + x - 6 = 0

➡ x² + (3x - 2x) - 6 = 0

➡ x² + 3x - 2x - 6 = 0

➡ x(x + 3) - 2(x + 3) = 0

➡ (x + 3) (x - 2)

➡ x = -3, x = 2

therefore the value of :-

alpha = -3

beta = 2

hence, 1/alpha + 1/beta = -1/3 + 1/2

taking LCM of 3 and 2 = 3 × 2 = 6

= (-1 × 2)/(3 × 2) + (1 × 3)/(2 × 3)

= -2/6 + 3/6

= 1/6

☆☞the value of 1/α + 1/ß is = 1/6

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$<marquee> Thank You </marquee>$

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Answer 2

Step-by-step explanation:

Given : α and β are the zeroes of the quadratic polynomial f(x)= x² - x - 4

On comparing with ax² + bx + c,

a = 1 , b= -1 , c= -4

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 = -4/1 = - 4

αβ = - 4 ……………………(2)

1/α + 1/β  - αβ = [( α+β) / αβ] - αβ

By Substituting the value from eq 1 & eq2 , we get  

= [ 1/−4 ]  - (- 4)

= −1/4 + 4

= (− 1 + 16)/4

=  15/ 4

1/α + 1/β  - αβ  = 15/4

Hence, the value of  1/α + 1/β  - αβ  = 15/4