# How do you solve the quadratic equation by completing the square: 2x^2 - 7x = 2?

Jul 14, 2015

The solution is $x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$.

#### Explanation:

$2 \left({x}^{2} - \frac{7}{2} x\right) = 2$
${x}^{2} - \frac{7}{2} x + \frac{49}{16} = 1 + \frac{49}{16}$
${\left(x - \frac{7}{4}\right)}^{2} = \frac{65}{16}$
$\left(x - \frac{7}{4}\right) = \pm \frac{\sqrt{65}}{4}$

$x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$

Jul 15, 2015

$x = \frac{7 + \sqrt{65}}{4} ,$ $\frac{7 - \sqrt{65}}{4}$

#### Explanation:

$2 {x}^{2} - 7 x = 2$

Divide both sides by $2$.

${x}^{2} - \frac{7}{2} x = 1$

To complete the square means to force a perfect square trinomial on the left side of the equation in the form ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$.

Divide the coefficient of the $x$ term by $2$, square the result, and add to both sides of the equation.

$\frac{- 7}{2} \div 2 = \left(- \frac{7}{2}\right) \cdot \frac{1}{2} = - \frac{7}{4}$
${\left(- \frac{7}{4}\right)}^{2} = \frac{49}{16}$

${x}^{2} - \frac{7}{2} x + \frac{49}{16} = 1 + \frac{49}{16}$

The common denominator for $1$ and $\frac{49}{16}$ is $16$. Multiply $1$ times $\frac{16}{16}$, then add the two fractions.

${x}^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{16}{16} + \frac{49}{16}$ =

${x}^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{65}{16}$

We now have a perfect square trinomial on the left side, where $a = x$ and $b = \frac{7}{4}$.

${\left(x - \frac{7}{4}\right)}^{2} = \frac{65}{15}$

Take the square root of both sides.

$x - \frac{7}{4} = \pm \sqrt{\frac{65}{16}}$ =

$x - \frac{7}{4} = \pm \frac{\sqrt{65}}{4}$

Solve for $x$.

$x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$ =

$x = \frac{7 \pm \sqrt{65}}{4}$

$x = \frac{7 + \sqrt{65}}{4}$ =

$x = \frac{7 - \sqrt{65}}{4}$