How do you solve the quadratic equation by completing the square: y^2+3y=10?

Apr 22, 2018

$y = 2 \mathmr{and} y = - 5$

Explanation:

${y}^{2} + 3 y \text{ } = 10$

You want to add in a missing term on the left so that you have a perfect square trinomial which can then be written in the form
(y +?)^2 This process is called 'completing the square.'

The 'missing term' is found from ${\left(\frac{b}{2}\right)}^{2}$ and has to be added to both sides of the equation.

${y}^{2} + 3 y \text{ "color(blue)( +(3/2)^2) = 10" } \textcolor{b l u e}{+ {\left(\frac{3}{2}\right)}^{2}}$

The left side can now be written as (y+?)^2 and the right side can be simplified.

${\left(y + \frac{3}{2}\right)}^{2} = 10 + \frac{9}{4} \text{ } \leftarrow$ square root both sides

$\left(y + \frac{3}{2}\right) = \pm \sqrt{10 + 2 \frac{1}{4}} = \pm \sqrt{12 \frac{1}{4}}$

$y + \frac{3}{2} = \pm \sqrt{\frac{49}{4}}$

$y = \frac{7}{2} - \frac{3}{2} \text{ "or" } y = - \frac{7}{2} - \frac{3}{2}$

$y = \frac{4}{2} = 2 \text{ "or " } y = - \frac{10}{2} = - 5$