Question

If
$\alpha \: and \: \beta \: are \: zeroes \: of \: polynomial \: p(x) = x { - mx + n \: then \: find \: values \: of \: \alpha \div \beta + \beta \div \alpha }^{2}$
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Answer 1

α and β are zeros of p(x)= x- mx+n.

so, they are the roots of : p(x)=0.

Then, we have , α-mα+n=0 .

or, α(1-m) + n=0.

or, α= - n/(1-m) = n/(m -1).

similarly, for β , we have, β= n/(m -1).

Therefore, α=β= n/(m -1) and 1/α= 1/β = (m- 1)/n.

Now, (α/β )+{ α/(β×β)} = 1+(1/β)= 1+ {(m -1)/n}= (m+n- 1)/n.

Answer 2

Question:

If α and ß are the zeros of the quadratic polynomial p(x) = x² - mx + n , then find the value of 1/α + 1/ß .

Answer:

1/α + 1/ß = m/n

Note:

★ The possible values of the variable for which the polynomial becomes zero are the zeros of the polynomial.

★ To find the zeros of the given polynomial , equate it to zero.

★ A quadratic polynomial can have atmost two 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 , (αß) = c/a

Solution:

Here,

The given quadratic polynomial is ;

p(x) = x² - mx + n

Clearly,

a = 1

b = -m

c = n

Thus,

=> Sum of zeros = -b/a

=> α + ß = -(-m)/1 = m

Also,

=> Product of zeros = c/a

=> αß = n/1 = n

Now,

1/α + 1/ß = (ß + α) / αß

= (α + ß) / αß

= m/n

Hence,

The required value of 1/α + 1/ß is m/n .