Question

If zeroes of polynomial x²-kx+2 are differ by 1 then value of k is

Answer 1

this is the answer of your questions

Answer 2

Concept:

If the zeros or solution of the quadratic equation ax² + bx + c = 0 are A and B then, the relation between coefficients and zeros are,

A + B = -b/a

AB = c/a

Given:

Given that, the zeros of polynomial x² - kx + 2 are differ by 1.

Find:

We have to find the value of k.

Solution:

Given that the quadratic equation is,

x²-kx+2 = 0

Comparing with ax² + bx + c = 0 we get,

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

Given that, the zeros of polynomial are differ by 1.

Let the zeros be A, A+1.

So, from the relations of zeros and coefficient we get,

A+A+1 = -(-k)/1

2A+1 = k

2A = k-1

k = 2A+1......(1)

and

A(A+1) = 2

A²+A -2 = 0

A²+2A-A-2 = 0

A(A+2)-1(A+2) = 0

(A+2)(A-1) = 0

A = 1,-2

So from (1) we get,

k = 2*1+1 = 2+1 = 3

k = 2(-2)+1 = -4+1 = -3

Hence the value of k are given by -3, 3.

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