# How do you solve the equation 4x^2+9x-1=0 by completing the square?

Jun 3, 2017

#### Answer:

Solution: $x = \frac{1}{8} \left(- 9 + \sqrt{97}\right) \mathmr{and} x = \frac{1}{8} \left(- 9 - \sqrt{97}\right)$

#### Explanation:

$4 {x}^{2} + 9 x - 1 = 4 \left({x}^{2} + \frac{9}{4} x\right) - 1 = 4 \left({x}^{2} + \frac{9}{4} x + {\left(\frac{9}{8}\right)}^{2}\right) - \frac{81}{16} - 1$ or

$4 {\left(x + \frac{9}{8}\right)}^{2} = \frac{81}{16} + 1 = 4 {\left(x + \frac{9}{8}\right)}^{2} = \frac{97}{16} = {\left(x + \frac{9}{8}\right)}^{2} = \frac{97}{64}$ or

$\left(x + \frac{9}{8}\right) = \pm \sqrt{\frac{97}{64}} \mathmr{and} \left(x + \frac{9}{8}\right) = \pm \frac{\sqrt{97}}{8}$ or

$\therefore x = - \frac{9}{8} \pm \frac{\sqrt{97}}{8} \mathmr{and} x = \frac{1}{8} \left(- 9 \pm \sqrt{97}\right)$

Solution: $x = \frac{1}{8} \left(- 9 + \sqrt{97}\right) \mathmr{and} x = \frac{1}{8} \left(- 9 - \sqrt{97}\right)$ [Ans]

Jun 3, 2017

#### Answer:

$x = - \frac{9}{8} \pm \frac{\sqrt{97}}{8}$

#### Explanation:

Step 1. Factor out the 4 from in front of the ${x}^{2}$

$4 \left({x}^{2} + \frac{9}{4} x - \frac{1}{4}\right) = 0$

Step 2. Take the coefficient of the $x$ term (i.e., $9 / 4$), cut it in half, square it, and add it and subtract it inside the parenthesis.

When you add a number and subtract it, that's like adding $0$ and so doesn't change the problem.

Middle term: $\frac{9}{4}$

Cut it in half: $\frac{9}{8}$ (this number will be used in Step 3)

Square the result: ${\left(\frac{9}{8}\right)}^{2}$

Next, add and subtract this term

$4 \left({x}^{2} + \frac{9}{4} x \text{ } \textcolor{red}{+ {\left(\frac{9}{8}\right)}^{2}} \textcolor{red}{- {\left(\frac{9}{8}\right)}^{2}} - \frac{1}{4}\right) = 0$

Step 3. Use the left three terms to make a perfect square

$4 \left(\textcolor{b l u e}{{x}^{2} + \frac{9}{4} x + {\left(\frac{9}{8}\right)}^{2}} - {\left(\frac{9}{8}\right)}^{2} - \frac{1}{4}\right) = 0$

$4 \left(\textcolor{b l u e}{{\left(x + \frac{9}{8}\right)}^{2}} - {\left(\frac{9}{8}\right)}^{2} - \frac{1}{4}\right) = 0$

Step 4. Simplify the fraction and multiply $4$ back through.

$4 \left({\left(x + \frac{9}{8}\right)}^{2} - {\left(\frac{9}{8}\right)}^{2} - \frac{1}{4}\right) = 0$

$4 \left({\left(x + \frac{9}{8}\right)}^{2} - \frac{97}{64}\right) = 0$ (*Don't forget signs here like I had!)

$4 {\left(x + \frac{9}{8}\right)}^{2} - \frac{97}{16} = 0$

Step 5. Use algebra to solve for $x$

$4 {\left(x + \frac{9}{8}\right)}^{2} = \frac{97}{16}$

${\left(x + \frac{9}{8}\right)}^{2} = \frac{97}{64}$

$x + \frac{9}{8} = \pm \sqrt{\frac{97}{64}}$

$x = - \frac{9}{8} \pm \frac{\sqrt{97}}{8}$