How do you solve the quadratic equation by completing the square: $y^{2} + 16 y = 2$ $y^2 + 16y = 2$ ?

Question

2521 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T17:57:09

$y = - 8 \pm \sqrt{66}$ $y= -8+-sqrt(66)$

to complete the square you use the formula:

$a x^{2} + b x + c$ $ax^2+bx+c$

a must equal 1

$c = {(\frac{b}{2})}^{2}$ $c=(b/2)^2$

the completed square is:

${(x + \frac{b}{2})}^{2}$ $(x+b/2)^2$

Here we go, in your function the y is the general formula's x:

$y^{2} + 16 y = 2$ $y^2 + 16y = 2$

$y^2 + 16y +underbrace(c = 2+c)$
we add c to both sides so we don't alter the equation

now solve c:

$c=(b/2)^2 = (16/2)^2=64$

$y^2 + 16y +64 = 2+64$

now complete the square:

$(y+8)^2 = 66$

Now solve:

$sqrt((y+8)^2) = +-sqrt(66)$

$y+8 = +-sqrt(66)$

$y= -8+-sqrt(66)$

How do you solve the quadratic equation by completing the square: y^2 + 16y = 2y2+16y=2y^2 + 16y = 2?