Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $f\left(x\right)=x^2-5x+k$ such that $\alpha -\beta =1$ find the value of $k$

• A. $k=24$
• B. $k=6$
• C. $k=-24$
• D. $k=-6$

Answer 1

The correct option is B $k=6$

Given $f\left(x\right)=x^2-5x+k$

and $\alpha -\beta ...\left(1\right)$

where $\alpha$ and $\beta$ are zero's of polynomial

Therefore, $f\left(x\right)=x^2-5x+k=0$

Express in terms of $a{x}^2+bx+c=0$

where, $a=1$

$b=-5$

$c=k$

and $\alpha +\beta =-\left(\frac{b}{b}\right)=\mathrm{5....}\left(2\right)$

$\alpha \beta =\frac{c}{a}=k....\left(3\right)$

squaring equation $\left(1\right)$ both sides

$(\alpha -\beta {)}^2=(1{)}^2$

$({\alpha }^2+{\beta }^2+2\alpha \beta )-2\alpha \beta -2\alpha \beta =1$

$(\alpha +\beta {)}^2-4\alpha \beta =1$

putting values from equation $\left(2\right)$ and $\left(3\right)$

$(5{)}^2-4k=1$

$25-4k=1$

$4k=24$

$k=6$