Question

If $\alpha and\beta$ are the zeroes of the quadratic polynomial $f\left(x\right)=K{x}^2+2x-15$ such that ${\alpha }^2+{\beta }^2=34.$ Find the value of K.

• A.
  
  $1and-\frac{2}{7}$
  
• B.
  
  $2and-\frac{2}{17}$
  
• C.
  
  $1and\frac{2}{17}$
  
• D.
  $1and-\frac{2}{17}$
  

Answer 1

The correct option is D

$1and-\frac{2}{17}$

$f\left(x\right)=K{x}^2+2x-15\alpha +\beta =-\frac{b}{a}=\frac{-2}{k}\alpha \beta =\frac{c}{a}=\frac{-15}{k}And,{\alpha }^2+{\beta }^2=34\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta -2\alpha \beta =34\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta =34(\alpha +\beta {)}^2-2\alpha \beta =34\Rightarrow {\left(\frac{-2}{k}\right)}^2-2\left(\frac{-15}{k}\right)=34\Rightarrow \frac{4}{k^2}+\frac{30}{k}=34\Rightarrow \frac{4+30k}{k^2}=34\Rightarrow 4+30k=34k^2\Rightarrow 0=34k^2-30k-4\Rightarrow 34k^2-30k-4=0\Rightarrow 34k^2(34-4)k-4=0\Rightarrow 34k^2-34k+4k-4=0\Rightarrow 34k(k-1)+4(k-1)=0\Rightarrow (k-1)(34k+4)=0k-1=0\therefore k=134k+4=0\Rightarrow K=\frac{-4}{34}\therefore K=\frac{-2}{17}$
Hence, the correct option is d.