2a^2 + 4a + 5 = 0 #solve the quadratic equation by completing the square and applying the square root property

Question

2a^2 + 4a + 5 = 0 #solve the quadratic equation by completing the square and applying the square root property 

Answer 1

Greetings fellow Buckeye. Originally from Youngstown,OH.

 

2a^2 + 4a +5 = 0

 

2a^2 + 4a      = -5   <--  subtracts 5 from both sides

 

2(a^2 + 2a       )  =  -5   <-- factors out 2 from left side

 

2(a^2 + 2a  +  1)    = -5 +2  <--- completing the square....

 

  2(   a      +    1)( a     +        1  ) = -3       <--- simplifies right side after factoring

 

2( a + 1)^2 = -3                 <--- rewrites the factors with exponents

 

(a+1)^2 = -3/2         <-- divides both sides by 2

 

 square-root ( (a+1)^2)  = square-root (-3/2)  <--- takes the square root of both sides

 

The left side is simply a+1

 

 a+1 = +/-square-root(-3/2)  <--- taking the square root of both sides, the plus and minus signs are required;

                                                     I shall denote it by and write it as +/-, which is read "plus or minus"

 

a = -1  +/- square-root(-3/2)   <-- subtracts 1 from both sides. Writes it as -1 in the front of the right side

    =  +/- square-root(-3/2)-1    <-- or you can write it as "subtracting 1";   adding -1 is the same as subtracting 1

 

Concentrating and focusing on just the part that says square-root(-3/2).....

 

square-root(-3/2) = square-root(-1* 3/2)                           <-- factors -1 out 

                              = square-root(-1)*square-root(3/2)     <-- property of square-root, namely 

                                                                                                  square-root(A*B) = square-root(A)*square-root(B)

 

                              = square-root(-1)* (square-root( (3*2)/(2*2))  <--- rationalizes the denominator by multiplying

                                                                                                              top and bottom by 2

 

                              = square-root(-1) * square-root( 6/4 ) 

 

                              = square-root(-1) * square-root(6)/square-root(4)   <--- another property of square roots. This time,

                                                                                                              square-root(A/B) = square-root(A)/square-root(B)

 

                              = square-root(-1) * rad6/2    <--- square-root(6) = rad6  ; slang term for "radical 6" or simply

                                                                                                                          the square root of six

 

                               = i * rad6/2         <--- square root of -1 is the imaginary number i

                                                               

 

So putting all of these pieces together, the final answer is:

 

 a = -1 +/- i*rad6/2 = i*rad6/2 - 1

which is read as:

 

 variable A equals negative one plus or minus  i times the square-root of 6 over 2    Or

 variable A equals i times the square-root of 6 over 2 minus one.

 

Again, many people will say the slang term, "rad 6" instead of "the square root of 6"

 

Now we have to CHECK this by plugging it into the original equation.

 

For the positive case, A = i*rad6/2 - 1.

Squaring this (using FOIL) gives:

 

          (i*rad6/2 - 1)^2 =  (i*rad6/2 - 1)  (i*rad6/2 - 1)

 

         [i*rad6/2]^2 = i*rad6/2*i*rad6/2 = i*i*rad6/2*rad6/2 = -1*6/4 = -3/2 <--- FIRST

         -i*rad6 /2  <--- OUTSIDE

         -i*rad6/2   <--- INSIDE

          +1 <--- LAST 

 

So the square of this complex number is : -3/2 - i*rad6 + 1  <--- notice the 2 cancels when combining the cross-terms  

                                                                 -1/2 - i*rad6 

  

doubling per the original quadratic equation gives 2(-1/2 - i*rad6/2) = -1 - 2i*rad6.

The quadratic term is complete and highlighted in bold.

 

The linear term is 4A = 4( i*rad6/2 - 1) = 2i*rad6 - 4 

 

So the left hand side is  -1 - 2i*rad6 + 2i*rad6 -4 + 5 = -1 -4 + 5           <-- 2i*rad6 cancels

                                                                                   = -5 + 5

                                                                                    = 0

 

This verifies that the solution A= i*rad6/2 - 1 is correct

 

The same check should be done with the negative case: A = -i*rad6/2 - 1.

We can be confident that it will check out, because these complex number

solutions MUST appear in conjugate pairs. I will leave this check for you.

If you do, you will see a lot of the same algebra, any terms will be the

same while some will change signs in just the right places. It will work. 

                       

It is a very complicated and challenging problem with lots of different things going on:

 FOIL method, operations with imaginary numbers, rationalizing denominators,

 operations with square roots,.... ALL of this on top of.... COMPLETING THE SQUARE.

 

Read, read, and re-read until it sinks in.... good luck.

 

Answer 2

Hi Lauren,

 

The procedure to complete the square is as follows:

1) Use addition and subtraction to move the constant term to the right and all other terms to the left.
2) Divide each term in the equation by the coefficient of the x 2 term, unless the coefficient is 1.
3) Determine the coefficient of the x term, divide it by two, square it, and add to both sides.
4) Factor the left side as a perfect square trinomial.
5) Take the square root of each side,

2a^2 + 4a = -5

a^2 + 2a = -5/2

a^2 + 2a + (2/2)^1 = -5/2 + (2/2)^1

(a + 1)^2 = -3/2

sqrt((a+1)^2) = sqrt(-3/2)

The solution will be complex number.