How do you "Solve this equation by completing the square"?

Question

The equation is :
ax2+bx+c=0 (where a>0)

And the steps are..
Step 1) axSquared+bx=-c
Step 2) xSquared +(b/a)x=-c/a
Step 3) ?????????

How do you figure this out, and what goes in Step 3.

Answer 1

The next step is adding b^2/4a^2 to both sides of the equation. This is half the coefficient of x, squared.

x^2 + (b/a)x + b^2/4a^2 = b^2/4a^2 - c/a

THe left side is a perfect square so

(x+b/2a)^2 = (b^2 - 4ac)/4a^2

Take square roots:

x+b/2a = +/- sqrt( b^2 - 4ac) / 2a

isolate x:

x= (b+/- sqrt(b^2-4ac)) / 2a (answer)

Answer 2

Ok jenny, a good question

Step 1)
write down equation, say: ax^2 + bx +c = 0
Step 2)
make the coefficient of x as unity(1) by dividing through out by a, which will make it
x^2 + (b/a)x + (c/a) = 0
Step 3)
Now add and subtract square of half the coefficient of x which is (b/a), which will make it
X^2 + (b/a)x + (1/2 b/a)^2 - (1/2 b/a)^2 + c/a = 0
Now this is of the form x^2 + 2xy + y^2 = (x+y)^2,
where x = x
y = 1/2 (b/a)
there equation becomes
(x + (1/2)(b/a))^2 - (1/2 b/a)^2 + c/a = 0
= (x + (1/2)(b/a))^2 = (1/2 b/a)^2 - c/a
thats it!! now solve

Answer 3

ill use ^2 to show squared
what you do is complete the square by adding on terms after it so that a square can be used on the left hand side of the equation
(x+b/2a)^2 = x^2 +b/ax +b^2/(4(a^2))
unfortunately you don't want the b^2/4(a^2) so you add it to the right hand side to leave both sides equal
(x+b/2a)^2 = -c/a + b^2/4a^2
now you can continue by taking a the square root of both sides (remembering that a root can be both negative and positive)
x+b/2a = +/- root of{(b^2/4a^2)-c/a}
if your looking for a numerical answer getting x should be easy if you want to do something theoretical then you'll probably need to re arrange it into the form
x= (-b+/-root[ b^2 -4ac])/2a

Answer 4

Once with numbers, once with variables.

3x² + 5x + 2 = 0
x² + (5/3)x = -2/3
x² + (5/3)x + (5/6)² = -2/3 + 25/36
(x + 5/6)² = (-24 + 25)/36
x + 5/6 = ±√( 1/36)
x = -5/6 ± 1/6

ax² + bx + c = 0
x² + (b/a)x = -c/a
x² + (b/a)x + [b/(2a)]² = -c/a + b²/(4a²)
[x + b/(2a)]² = -4ac/(4a²) + b²/(4a²)
x + b/(2a) = ±√[ (b² - 4ac)/(4a²)]
x = -b/(2a) ±√(b² - 4ac) / (2a)

Answer 5

Step 3 is ((1/2)*(b/a))^2. This answer is added to both sides
x^2+(b/a)x*+(b^2/4a^2)=-c/a+b/a
(x+b/2a)^2-(b^2-4ac)/4a^2
x+b/2a=+/- sqrt (b^2-4ac)/2a
x=(-b+/- sqrt(b^2-4ac))/2a
That's the Pythagorean theorem!

Answer 6

ax^2 + bx + c = 0
ax^2 + bx = -c
x^2 + (b/a)x = -c/a
x^2 + (b/a)(1/2)x + ((b/a)(1/2))^2 = -c/a + (b/a)(1/2)
x^2 + (b/2a)x + (b/2a)^2 = (-c/a) + (b/2a)

Answer 7

Divide (b/a) by two and square it, add to both sides of the equation.