Question

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2-3x+k$, such that $\alpha -\beta =1$, find the value of $k$.

Answer 1

Compare the given quadratic with the standard form.

We know that the quadratic equation in the standard form is given by, $a{x}^2+bx+c$

Given: $x^2-3x+k$

Comparing with the standard form of quadratic equation.

$a=1,b=-3,c=k$

Since, Sum of zeroes, $\alpha +\beta =\frac{-b}{a}=3$

Product of zeroes, $\alpha \beta =\frac{c}{a}=k$

Now,

$$\begin{gathered} {(\alpha +\beta )}^2-{(\alpha -\beta )}^2=4\alpha \beta ;\mathbf{}\left[\right(x+y)²–(x–y)²=4xy] \\ \Rightarrow {\left(3\right)}^2-{\left(1\right)}^2=4k \\ \Rightarrow 4k=9-1 \\ \Rightarrow 4k=8 \\ \Rightarrow k=2 \end{gathered}$$

Hence, the value of $k$ is $2$.