How do you solve $y^{2} - 3 y - 14 = 0$ $y^2 - 3y - 14 = 0$ using completing the square?

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Answer 1 · 2015-07-06T12:03:52

Add 14 to both sides, then add half the square of the coefficient of $y$ $y$ to both sides, then solve.

1) Add 14 to both sides of the equation

$y^{2} - 3 y = 14$ $y^2-3y=14$

2) Add half the square of $y$ $y$ 's coefficient to both sides

$y^{2} - 3 y + {(- \frac{3}{2})}^{2} = 14 + {(- \frac{3}{2})}^{2}$ $y^2-3y + (-3/2)^2 = 14 + (-3/2)^2$

3) Now the left side is a perfect square; remember to simplify the right hand side too

${(y - \frac{3}{2})}^{2} = \frac{65}{4}$ $(y-3/2)^2=65/4$

4) Take $\pm$ $+-$ the square root of both sides; remember the left side is a square, so it remains positive (i.e., it's not $\pm$ $+-$ )

$y - \frac{3}{2} = \pm \sqrt{\frac{65}{4}}$ $y-3/2=+-sqrt(65/4)$

5) Add the remaining fraction on the left to both sides and simplify your roots

$y = \frac{3}{2} \pm \frac{\sqrt{65}}{2}$ $y=3/2+-sqrt(65)/2$

The solution set is: $[\frac{3}{2} - \frac{\sqrt{65}}{2}, \frac{3}{2} + \frac{\sqrt{65}}{2}]$ $[3/2-sqrt(65)/2, 3/2+sqrt(65)/2]$

How do you solve y2−3y−14=0 y^2 - 3y - 14 = 0 using completing the square?