How can i solve the quadratic equation x2 - 4x + 57 = -5 by completing the square?

Question

johnweldon1993 · Answer

First things first...you want it in the form

$$\large ax^2 + bx = c$$

so subtract 57 from both sides of the equation...what do we have now?

anonymous · Answer

x2 - 4x = -5 ?

anonymous · Answer

@johnweldon1993

johnweldon1993 · Answer

Remember you have to subtract it from the other side too...

$$\large x^2 - 4x + 57 - 57 = -5 - 57$$

what do you have after that?

anonymous · Answer

dont know /: @johnweldon1993

johnweldon1993 · Answer

Don't get discouraged...*ps...this is going to be a complex answer I can already tell...

$$\large x^2 - 4x + 57 - 57 = -5 - 57$$

Just simplify here.... + 57 - 57 = 0....and -5 - 57 = -62...so now we have

$$\large x^2 - 4x = -62$$

okay?

anonymous · Answer

Just wrote that down, thank you! @johnweldon1993

johnweldon1993 · Answer

We are not even CLOSE to done I hope you know that lol...

johnweldon1993 · Answer

That was just getting it into the right form to do the procedure....now we begin :D ready?

anonymous · Answer

Oh i thought we were lol awks. Yes :)
@johnweldon1993

johnweldon1993 · Answer

Nope :P lol

okay...now you have an equation in the form

$$\large ax^2 + bx = c$$

*compare it to your equation...

$$\large x^2 - 4x = -62$$

look the same?

anonymous · Answer

Yes thats the standard form of a quadratic functio, that's what i was trying to find lol @johnweldon1993 totally forgot how it was formed

anonymous · Answer

*function

johnweldon1993 · Answer

Perfect....now have you done 'completing the square' before?

anonymous · Answer

Can you refresh my mind on what it is again? @johnweldon1993

johnweldon1993 · Answer

Absolutely :)

So you get you equation into standard form *check* lol

Now you take your 'b' value.....divide it by 2....then square the result...what is this?

anonymous · Answer

2?
@johnweldon1993

johnweldon1993 · Answer

Not quite...you didnt square it...

your 'b' value is -4 right ....you divide that by 2.....-4/2 = -2.......now you square that result....-2² = 4....right?

anonymous · Answer

Oh yes, my mistake @johnweldon1993

johnweldon1993 · Answer

No problem...so now...you add that to both sides of the equation...

$$\large x^2 - 4x + 4 = -62 + 4$$

simplified down a bit we have

$$\large x^2 - 4x + 4 = -58 $$

Now you would factor the left hand side...but...remember what I said...this is going to have imaginary results...*complex numbers*

can you do that?

johnweldon1993 · Answer

We know we are looking for a perfect square here....so what number times itself...= 4....but also when added to itself = -4?

anonymous · Answer

-2 @johnweldon1993

johnweldon1993 · Answer

Right...so we have

$$\large (x - 2)^2 = -58$$

Now take the square root of both sides *this is where the complex numbers comes in* lol...and I'll brb have an errand to run :)

johnweldon1993 · Answer

I hope you got this @justagal1619

From the step before....take the aquare root of (x - 2)^2 = -58

this becomes

$$\large \sqrt{(x - 2)^2} = \sqrt{-58}$$

Which simplifies into

$$\large (x - 2) = \pm \sqrt{-58}$$

Now when you solve for 'x'...you just add 2 to both sides of the equation...

$$x = 2 \pm \sqrt{-58}  $$

anonymous · Answer

@johnweldon1993 kinda thanks anyways :-)