which value must be added to the expression $x^{2}-3x$ to make it a perfect - square trinomial?\n$\frac{3}{2}…

which value must be added to the expression $x^{2}-3x$ to make it a perfect - square trinomial?\n$\frac{3}{2}$\n$\frac{9}{4}$\n$6$\n$9$

which value must be added to the expression $x^{2}-3x$ to make it a perfect - square trinomial?\n$\frac{3}{2}$\n$\frac{9}{4}$\n$6$\n$9$

Answer

Explanation:

Step1: Recall the formula for perfect - square trinomial

For a quadratic expression of the form $x^{2}+bx$, to make it a perfect - square trinomial, we add $\left(\frac{b}{2}\right)^{2}$. In the given expression $x^{2}-3x$, we have $b = - 3$.

Step2: Calculate the value to be added

Substitute $b=-3$ into $\left(\frac{b}{2}\right)^{2}$. We get $\left(\frac{-3}{2}\right)^{2}=\frac{(-3)^{2}}{2^{2}}=\frac{9}{4}$.

Answer:

B. $\frac{9}{4}$