Question

s) If alpha,beta are roots of x ²-k(x+1)-c=0 then (I+a) (1+B)=​

Answer 1

$First, we have the equation: x2−a(x+1)−b=0x2−a(x+1)−b=0</p><p>We’ll rewrite this in the ‘standard’ format: x2−ax−(a+b)=0x2−ax−(a+b)=0 [Eq. 1]</p><p>Next, as p and q are the roots of this quadratic, we have: (x−p)(x−q)=0(x−p)(x−q)=0</p><p>Expanding the left side: x2−(p+q)x+pq=0x2−(p+q)x+pq=0 [Eq.2]</p><p>But equations 1. & 2. must be the same; therefore, equating coefficients, we have:</p><p>p+q=ap+q=a and pq=−(a+b)pq=−(a+b)</p><p>Finally, you want me to evaluate (p+1)(q+1)(p+1)(q+1)</p><p>=pq+p+q+1=pq+p+q+1</p><p>=−(a+b)+a+1=−(a+b)+a+1</p><p>=1−b</p><p>$