Question

How do you solve $2x^2-7x+3=0$ by completing the square.

Answer 1

Compute the roots of the quadratic equation.

Given: $2x^2-7x+3=0$

$\Rightarrow$ $2(x^2-\frac{7}{2}x)+3=0$

Making a perfect square by adding and subtracting $2{\left(\frac{7}{4}\right)}^2$

$2\left[x^2-\frac{7}{2}x+{\left(\frac{7}{4}\right)}^2\right]-2{\left(\frac{7}{4}\right)}^2+3=0$

$\Rightarrow$ $2{\left[x-\frac{7}{4}\right]}^2-\frac{49}{8}+3=0$

$\Rightarrow$ $2{\left[x-\frac{7}{4}\right]}^2=\frac{25}{8}$

$\Rightarrow$ ${\left[x-\frac{7}{4}\right]}^2=\frac{25}{16}$

$\Rightarrow$ $\left[x-\frac{7}{4}\right]=\pm \frac{5}{4}$

$\Rightarrow$ $x=\pm \frac{5}{4}+\frac{7}{4}$

$\Rightarrow$ $x=3or\frac{1}{2}$

Hence, the roots of $2x^2-7x+3=0$ are $3$ and $\frac{1}{2}.$