To convert the standard form to vertex form, one needs a process called
completing the square.
Completing the Square
Say we have a simple expression like x2 + bx. Having x twice in the same expression can make life hard. What
can we do? Well, with a little inspiration from Geometry we can convert it, like this:
As you can see x2 + bx can be rearranged nearly into a square ...
... and we can complete the square with (b/2)2
In Algebra it looks like this:
x2 + bx
+ (b/2)2
"Complete the Square"
=
(x+b/2)2
So, by adding (b/2)2 we can complete the square.
And (x+b/2)2 has x only once, which is easier to use.
Keeping the Balance
Now ... you can't just add (b/2)2 without also subtracting it too! Otherwise the whole value would change.
So I will show you how to do it properly with an example:
Start with:
("b" is 6 in this case)
Complete the Square:
Also subtract the new term
Simplify it and we are done.
The result:
x2 + 6x + 7 = (x+3)2 - 2
From:
http://www.mathsisfun.com/alge
bra/completing-square.html
And now x only appears once, and your job is done!
Examples:
Completing the Square when you’re a coefficient (in standard form) is 1.
Completing the Square when you’re a coefficient (in standard form) is not 1.
Here are the steps for solving 2x2 + 8x +11
2x2 + 8x +11=0
Original equation set to 0 (set this to zero since the y value of 0 is where the xintercepts/solutions are found)
2
2x + 8x= -11
Subtract 11 from both sides
2
2(x + 4x)= -11
Factor out the 2 from the coefficients from the left side of the equation
2(x2 + 4x + 4) = -11+8
Divide the b coefficient by 2 and square it to find the c that makes perfect square
trinomial (4/2 = 2; 22=4). Since you added the 4 x2 to left side of the equation to
make it a perfect square trinomial, you need to add 4 x2 to the right side of the
equation also to keep the equation balanced
2
2(x+2) = -3
Simplify the right side of the equation, express the perfect square trinomial on the
left as a squared binomial
2 −3
Solve for x. First divide both sides by 2.
(x+2) = 2
Take the square root of both sides
−3
x+2=√
2
−3
𝑥 = −2√
2
Subtract 2 from both sides. These roots do not represent real numbers because the
number under the square roots sign is negative. Since every real number is on the
x-axis, this is not. Therefore, this equation does not have roots (x-intercepts) and so
has no solution.
Practice Problems