Question

If $\alpha ,\beta$ are the zeroes of $f\left(x\right)=x^2-5x+b$ and $\alpha -\beta =1$, then $b$ is

• A. 1
• B. 6
• C. 4
• D. 0

Answer 1

Answer: (B)

6

Given, $\alpha$ and $\beta$ are the zeros of $f\left(x\right)=x^2-5x+b$ and $\alpha -\beta =1$
Thus, sum of the zeros $=\alpha +\beta =5$ and product of the zeros $=\alpha \beta =b$
Now,
${(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta$
$=>{(\alpha -\beta )}^2={\left(5\right)}^2-4b$
$\Rightarrow {\left(1\right)}^2=25-4b$
$\Rightarrow 4b=24$
$\Rightarrow b=6$