Question

if alpha and beeta are the polynomial a^2-x-4x+k such that alpha - beeta = 9. find k

Answer 1

Answer:

Question:

If alpha and beta are the zeros of the polynomial x^2 - 4x + k such that

alpha - beta = 9. Find k=?

Solution:

Note:

If we a quadratic equation in variable x ,

say: ax^2 + bx + c , let alpha and beta are its zeros ,then;

alpha + beta = -b/a

alpha•bata = c/a

Here, the given quadratic polynomial is;

x^2 - 4x + k .

Clearly, here we have;

a = 1

b = - 4

c = k

It is given that, alpha and beta are the zeros of the given polynomial,

Thus;

alpha + beta = - b/a = -(-4)/1 = 4

alpha•beta = c/a = k/1 = k

Also, it is given that ,

alpha - beta = 9

Note:

(a+b)^2 = (a-b)^2 + 4ab.

We can use this identity to find the value of k.

Thus:

=> (alpha + beta)^2 = (alpha – beta)^2

+ 4•alpha•beta

=> (4)^2 = (9)^2 + 4k

=> 16 = 81 + 4k

=> 4k = 16 - 81

=> 4k = - 65

=> k = - 65/4

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

Alpha and beta are the zeroes of polynomial x^2-x-4x+k such that alpha - beta = 9.

To Find:

We need to find the value of k.

Solution:

The given polynomial is x^2 - 4x + k .

Comparing with ax^2 + bx + c we get,

a = 1

b = - 4

c = k

α + β = -b/a = -(-4)/1 = 4

αβ = c/a = k/1 = k

α - β = 9

We know (a+b)^2 = (a-b)^2 + 4ab.

or (α + β)^2 = (α - β)^2+ 4αβ

=> (4)^2 = (9)^2 + 4k

=> 16 = 81 + 4k

=> 4k = 16 - 81

=> 4k = - 65

=> k = - 65/4

Hence, the value of k is -65/4.