# How do you solve y^2-5y-3=0 by completing the square?

Apr 29, 2016

$y = \frac{5}{2} \pm \frac{\sqrt{37}}{2}$

#### Explanation:

We will also use the difference of squares identity, which can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 y - 5\right)$ and $b = \sqrt{37}$.

Pre-multiply by $4$ to reduce the amount of working in fractions:

$0 = 4 \left({y}^{2} - 5 y - 3\right)$

$= 4 {y}^{2} - 20 y - 12$

$= {\left(2 y - 5\right)}^{2} - 25 - 12$

$= {\left(2 y - 5\right)}^{2} - {\left(\sqrt{37}\right)}^{2}$

$= \left(\left(2 y - 5\right) - \sqrt{37}\right) \left(\left(2 y - 5\right) + \sqrt{37}\right)$

$= \left(2 y - 5 - \sqrt{37}\right) \left(2 y - 5 + \sqrt{37}\right)$

Hence:

$y = \frac{5}{2} \pm \frac{\sqrt{37}}{2}$

May 2, 2016

$y = 5.541$ OR $y = - 0.541$

#### Explanation:

${y}^{2} - 5 y = 3 \Rightarrow$ move the constant to the right hand side.

Complete the square by adding what is missing from the square of the binomial to BOTH sides. (b÷2)^2

${y}^{2} - 5 y + {\textcolor{red}{\frac{5}{2}}}^{2} = 3 + {\textcolor{red}{\frac{5}{2}}}^{2}$

${\left(y - \frac{5}{2}\right)}^{2} = 3 + \left(\frac{25}{4}\right)$

y - 5/2 = +-sqrt(9.25 ............$\Rightarrow 3 + 6 \frac{1}{4} = 9 \frac{1}{4}$

$y = \sqrt{9.25} + 2.5$ OR $y = - \sqrt{9.25} + 2.5$

$y = 5.541$ OR $y = - 0.541$