Question

if alpha and beta are the roots of the equation x^2-p(x+1)-q=0 then the value of alpha square +(2+p)alpha beta +beta square?

Answer 1

Originally Answered: If \alpha and \beta are the roots of the equation x^2 -p(x+1)-q=0 then the value of \frac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+q} + \frac{\beta^2+2\beta+1}{\beta^2+2\beta+q}is a) 1, B) 2, c) 3, d) 0?

The given equation can be written as  2−−(+)=0.  So,

+=  and  =−(+)

From here, we get  =−−−

Now the question asks us to find:

2+2+12+2++2+2+12+2+

So the expression gets reduced to:

2+2+12+2−−−+2+2+12+2−−−

=  (+1)2(+1)(−)+(+1)2(+1)(−)

=  (+1)(−)+(+1)(−)

=  (+1−−1)(−)

=  (−)(−)

=  1

So the solution to this problem is 1

Hope this helped!

Answer 2

Answer:

1

Step-by-step explanation:

If  α and  β  are the roots of the equation  x 2 − p ( x + 1 ) − q= 0  or  x 2 − p x − ( p + q ) = 0

then

α + β = p  and  α β = − ( p + q )

So, q

= − α − β − α β

Now the numerical value of  

α 2 + 2 α + 1 α 2 + 2 α + q + β 2 + 2 β + 1 β 2 + 2 β +q

= α 2 + 2 α + 1 α 2 + 2 α − α − β − α β + β2+2 β + 1 β 2 + 2 β − α − β − α β

= α 2 + 2α + 1 α 2 + α − β − α β + β 2 + 2 β + 1 β 2 + β −α − α β  

= ( α + 1 ) 2 α ( α + 1 ) − β ( α + 1 ) + ( β + 1 ) 2 β ( β + 1 ) − α ( 1 + β )  

= ( α + 1 ) 2 ( α − β ) ( α + 1 ) − ( β + 1 ) 2 ( β + 1 ) ( α − β )

 = α + 1 α − β − β + 1 α − β

 = α + 1 − β − 1α −β

= 1

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