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If \alpha , \beta be roots of x^{2}+px+1=0 and \gamma,\: \delta are the roots of x^{2}+qx+1=0 , then (\alpha-\gamma ) (\beta-\gamma )(\alpha +\delta )(\beta +\delta )=

  • Option 1)

    p^{2}+q^{2}

  • Option 2)

    p^{2}-q^{2}

  • Option 3)

    q^{2}-p^{2}

  • Option 4)

    None

 

Answers (1)

As we learnt in 

Sum of Roots in Quadratic Equation -

\alpha +\beta = \frac{-b}{a}

- wherein

\alpha \: and\beta are root of quadratic equation

ax^{2}+bx+c=0

a,b,c\in C

 

 

Product of Roots in Quadratic Equation -

\alpha \beta = \frac{c}{a}

- wherein

\alpha \: and\ \beta are roots of quadratic equation:

ax^{2}+bx+c=0

a,b,c\in C

 

 Using sum and product formula

\alpha +\beta =-p;\alpha \beta =1\\*\\*r+\delta =-q;r\delta =1\\*\\*\left (\alpha -r \right )\left (\beta +\delta \right )\left ( \beta -r \right )\left ( \alpha +\delta \right )\\*\\*= > \left (\alpha \beta +\alpha \delta -\beta r-r\delta \right )\left (\alpha \beta +\beta \delta -\alpha r-r\delta \right )\\*\\*= > \left (\alpha \delta -\beta r \right )\left ( \beta \delta -\alpha r \right )\\*\\*= > \alpha \beta \delta ^{2}-r\delta \alpha ^{2}-r\delta \beta ^{2}+\alpha \beta r^{2}\\*\\*r^{2}+\delta ^{2}-\left ( \alpha ^{2}+\beta ^{2} \right )\\*\\*= > \left ( r+\delta \right )^{2}-2\delta r-\left ( \alpha +\beta \right )^{2}+2\alpha \beta\\*\\*q^{2}-p^{2}


Option 1)

p^{2}+q^{2}

Incorrect

Option 2)

p^{2}-q^{2}

Incorrect

Option 3)

q^{2}-p^{2}

Correct

Option 4)

None

Incorrect

Posted by

Vakul

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