Question

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2-5x+m$ such that $\alpha -\beta =1$, then find the value of $m$.

Answer 1

We know that for a general quadratic equation $a{x}^2+bx+c$:

Sum of zeros $=\frac{-b}{a}$ and Product of zeros $=\frac{c}{a}$

$\therefore \alpha +\beta =\frac{-b}{a}=-\frac{(-5)}{1}=+5$

$\alpha +\beta =5...\left(i\right)$

$\alpha -\beta =1...\left(ii\right)$

Add equation (i) & (ii) we get,

$2\alpha =6$

It gives $\alpha =3$

Putting the above value in (1) we get,

$\beta =2$

$\to$ multiplication of two roots is $\frac{c}{a}.$

$\therefore \alpha .\beta =\frac{c}{a}$

$\therefore 3\times 2=\frac{m}{1}$

$\therefore m=6$