Math, asked by pavi3813, 7 months ago

If alpha and beta are the zeroes of the polynomial x^2-3x+k , such that alpha-beta=1, find the value of k

Answers

Answered by Anonymous
36

Question:

If α and ß are the zeros of the given polynomial

x² - 3x + k , such that α - ß = 1 , then find the value of k .

Answer:

k = 2

Note:

• A polynomial of degree 2 is called quadratic polynomial.

• The general form of a quadratic polynomial is given by ; ax² + bx + c = 0 .

• The possible values of variable ( or unknown ) for which the polynomial becomes zero are known as its zeros.

• If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

Sum of zeros , (α + ß) = -b/a

Product of zeros , α•ß = c/a

• If α and ß are the zeros of any quadratic polynomial then that polynomial is given as ;

x² - (α + ß)x + α•ß .

Solution:

Here ,

The given quadratic polynomial is ; x² - 3x + k

Clearly , we have ;

a = 1

b = -3

c = k

Now,

The sum of the zeros of the given quadratic polynomial is ;

=> α + ß = -b/a

=> α + ß = -(-3)/1

=> α + ß = 3 ------(1)

Also,

The product of zeros of the given quadratic polynomial is ;

=> α•ß = c/a

=> α•ß = k/1

=> α•ß = k -------(2)

Also,

It is given that ,

α - ß = 1 ------(3)

Now,

We know that ; (x + y)² - (x - y)² = 4xy

Thus,

=> (α + ß)² - (α - ß)² = 4αß

=> 3² - 1² = 4k

=> 4k = 3² - 1²

=> 4k = 9 - 1

=> 4k = 8

=> k = 8/4

=> k = 2

Hence,

The required value of k is 2 .

Answered by VishnuPriya2801
20

Answer:-

Given polynomial => x² - 3x + k = 0

 \alpha  -  \beta  = 1 \:  -  \: equation \: (1)

We know that,

sum \: of \: the \: roots \: ( \alpha  +  \beta ) =  \frac{ - b}{a}  \\  \\  \alpha  +  \beta  =  \frac{ - ( - 3)}{1}  \\  \\  \alpha  +  \beta  = 3 \:  -  \: equation \: (2)

Add both the equations,

2 \alpha  =  3 + 1 \\  \\ 2 \alpha  = 4 \\  \\  \alpha  =  \frac{4}{2}  = 2

substitute \:  \alpha  \: value \: in \: equation \: (1) \\  \\  \alpha  -  \beta  = 1 \\  \\ 2 -  \beta  = 1 \\  \\   \beta  = 1

Substitute , the roots in the given polynomial.

(2)² - 3(2) + k = 0

4 - 6 + k = 0

k = 2

(Or)

(1)² - 3(1) + k = 0

1 - 3 + k = 0

k = 2

Hence, the value of k is 2.

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