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Question

If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate :

(i) α − β
(ii) 1α-1β
(iii) 1α+1β-2αβ
(iv) α2β − αβ2
(v) α4 + β4
(vi) 1aα+b+1aβ+b
(vii) βaα+b+αaβ+b
(viii) aα2β+β2α+bαβ+βα

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Solution

(i) Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

Substituting and then we get,

Hence, the value of is.

(ii) Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

Substituting and then we get,

By taking least common factor we get,

Hence the value of is .

Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

By cross multiplication we get,

By substituting and we get ,

Hence the value of is.

Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

By taking common factor we get,

By substituting and we get ,

Hence the value of is.

Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

By substituting and we get ,

By taking least common factor we get

α4+β4=-b a2-2×ca2-2×ca2= b2a2-2ca2-2×ca2= b2-2aca22-2×c2a2= b2-2ac2a4-2×c2a2=b2-2ac2-2c2a2a4Hence the value of α4+β4 is b2-2ac2-2c2a2a4.

(vi) Since α and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

(vii) Since α and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

(viii) Since α and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .


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