Answer
Given: $x^2-3x$
The coefficient of the $x$-term is $-3$. Half of $-3$ is $-\frac{3}{2}$, and $(-\frac{3}{2})^2 = \frac{9}{4}$
Thus, add $\frac{9}{4}$ to both sides to complete the square.
$x^2-3x+\frac{9}{4}=\frac{9}{4}$
Since, $a^2+2ab+b^2=(a+b)^2$, the equation becomes $(x-\frac{3}{2})^2=\frac{9}{4}$
Apply the square root property to get the value of $x$.
$x-\frac{3}{2}=±\sqrt{\frac{9}{4}}$
$x=\frac{3}{2}±\frac{3}{2}$
$x=\frac{3}{2}+\frac{3}{2}$ or $x=\frac{3}{2}-\frac{3}{2}$
$x=3$ or $x=0$
Work Step by Step
Given: $x^2-3x$
The coefficient of the $x$-term is $-3$. Half of $-3$ is $-\frac{3}{2}$, and $(-\frac{3}{2})^2 = \frac{9}{4}$
Thus, add $\frac{9}{4}$ to both sides to complete the square.
$x^2-3x+\frac{9}{4}=\frac{9}{4}$
Since, $a^2+2ab+b^2=(a+b)^2$, the equation becomes $(x-\frac{3}{2})^2=\frac{9}{4}$
Apply the square root property to get the value of $x$.
$x-\frac{3}{2}=±\sqrt{\frac{9}{4}}$
$x=\frac{3}{2}±\frac{3}{2}$
$x=\frac{3}{2}+\frac{3}{2}$ or $x=\frac{3}{2}-\frac{3}{2}$
$x=3$ or $x=0$