completing the square x ^2 +6x -8 =0

Question

johnweldon1993 · Answer

You would first add 8 to both sides to get it in the correct form

$$\large x^2 + 6x = 8$$

You now have an equation in the form

$$\large ax^2 + bx = c$$

You take your 'b' number (6) divide it by 2...and square the result...and add that to both sides of the equation..what do you get when you do that?

anonymous · Answer

you get 9 and add to bith sides ?

johnweldon1993 · Answer

Right! you add 9 to both sides...so now we have

$$x^2 + 6x + 9 = 8 + 9$$

Next step....factor that left hand side of the equation...

anonymous · Answer

ooohh can you help me with that ..

johnweldon1993 · Answer

No problem..the thing is you'll see that you get a perfect square..

What 2 numbers multiply to get 9...and also add to get 6? 3 and 3 comes to mind

$$\large x^2 + 6x + 9 \space \text {can be factored into} \space (x+3)(x+3)$$

Now its dumb to write (x + 3) twice...so we write

$$(x + 3)^2 = 17$$

What would we do next?

anonymous · Answer

(X+ 3) ^2 - 17 = 0 ?

anonymous · Answer

is that the answer then ?

johnweldon1993 · Answer

No you wouldnt bring the 17 over...you would square root both sides of the equation

$$x + 3 = \pm \sqrt{17}$$

Now you would solve the 2 equations

anonymous · Answer

oh ok

johnweldon1993 · Answer

so

$$x + 3 = \sqrt{17}$$
and

$$x + 3 = -\sqrt{17}$$

Solve for each 'x'

anonymous · Answer

-3-sqrt(17)

anonymous · Answer

x = sqrt 17 - 3