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About Josh
Expertise
When I work through problems, I emphasize principles and key ideas which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponentials & Logarithms. Basic Calculus. Probability, Permutations and Combinations. Mathematical Induction. Complex numbers. Physics problems.

Experience

Experience:
I have worked as a teaching assistant in college. My hope is that more people will share knowledge without boundary, give help without seeking recognition or monetary rewards.

Supplementary Website:
See a selection of past questions in my maths repository under "Question Archive"

Education Credentials:
Bachelor degree in Engineering Science.

"Everyone struggles with something."

You are here: Experts > Science > Math for Kids > Basic Math > completing the square in quadratic equations

Basic Math - completing the square in quadratic equations

Expert: Josh - 11/4/2009

Question
Hi Josh,out of 50 questions,these three i had a problem with. Can you help me with solving for x,in steps please,

1.    10x^2+7x-12 =0

2.    x^2+4x-8    =0

3.    10+6x -x^2 =0

Thanks for your time

Answer
Hi Richard,

"Completing the square" may not be the easiest way of finding the answer to these quadratic equations. Before we go any further, remember that any complex looking quadratic equations can be solved using the quadratic equation. Without formal derivation, the quadratic formula states that for a*x^2 + b*x + c = 0 (any real-valued constants or numbers "a", "b" and "c"), the solutions (if they exist [see note 1]) are given by x = [-b + sqrt(b^2-4*a*c)]/(2a) and x = [-b - sqrt(b^2-4*a*c)]/(2a). Watch out for the brackets ok, we can always use this as a last resort to finding the answer.

| note 1: by this, I mean a real solution does not exist if the expression inside
| the square root (called the "discriminant") is negative.

The bottom line is unless the question asks you to solve a quadratic in a specific way, you can use any technique you like to find the solution(s).

For Q1, we can factorize 10 out of the equation for a start. This does not affect the solution in any way. Why?? Because we are effectively dividing both sides of the equation by 10. Performing the same arithmetic operation on both sides does not alter the solution. This, in fact, is the basis for solving simultaneous equations.
We then have x^2+0.7x-1.2=0. Next, we want to complete the square, so to speak. i.e., manipulating the first two terms into a perfect square (the trick is) without altering the equation.

(i) Observe that we can split x^2 + 0.7x into x^2 + 0.35x + 0.35x. If we have x^2 + 0.35x + 0.35x + (0.35*0.35), we would have a perfect square (x+0.35)^2. So, the constant "(0.35)^2" is the only thing missing.

(ii) Observe that x^2 + 0.35x + 0.35x + (0.35)^2 - (0.35)^2 is in perfect harmony with (identical to) x^2 + 0.7x (what we started with). So, we can rewrite x^2+0.7x-1.2=0 as
x^2 + 0.35x + 0.35x + (0.35)^2 - (0.35)^2 -1.2 = 0
| Take a moment to absorb this, go over the observations to see if it makes sense.
| As the first four terms constitute a perfect square (i.e., x^2+(b/2)x+(b/2)x+(b/2)^2
| = [x-(b/2)]^2, the whole thing is equivalent to
(x + 0.35)^2 - (0.35)^2 -1.2 = 0
Now the hard part is done....are we ok to press ahead?

(iii) We add (0.35)^2+1.2 to both sides of the equation.
This gives (x + 0.35)^2 = (0.35)^2 + 1.2

(iv) Take square root on both sides (with the plus and minus sign out the front)
x + 0.35 = +/- sqrt{(0.35)^2 + 1.2}

(v) Finally, subtract 0.35 from both sides, we end up with two possible solutions:
x = -0.35 + sqrt{(0.35)^2 + 1.2} and x = -0.35 -sqrt{(0.35)^2 + 1.2}

Comment: The procedure we have just completed is actually a validation of the quadratic formula.
Starting from 10x^2+7x-12 = 10(x^2+0.7x-1.2)=0, if we let a=1, b=0.7, c=1.2, we can clearly see that the answers we obtained (using completing the square) is identical to the quadratic formula x = [-b + sqrt(b^2-4*a*c)]/(2a) and x = [-b - sqrt(b^2-4*a*c)]/(2a).

Exercise: Verify this yourself. Put a=1, b=0.7, c=1.2, into x = [-b +/- sqrt(b^2-4*a*c)]/(2a). Are they identical to the roots of the quadratic equation?

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Q2. This is easier to deal with, since it involves only integer coefficients. You can retrace all the steps we have taken before. I won't provide commentary this time.
x^2 + 4x -8 = 0
x^2 + 2x + 2x -8 = 0
x^2 + 2x + 2x +(2)^2 -(2)^2 -8 = 0
(x+2)^2 -(2)^2 -8 = 0
(x+2)^2 = (2)^2 +8
x+2 = +/- sqrt{(2)^2 +8}
soln: x = -2 + sqrt{(2)^2 +8} and x = -2 - sqrt{(2)^2 +8}

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Q3. Use the same approach. Hint, multiply throughout by -1 may make things a little easier.

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