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Find the solutions of the equation x^2+2x+5=0. Use the method of completing the square.

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hahaz | Student, Grade 10 | eNoter

Posted June 18, 2010 at 2:52 AM via web

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Find the solutions of the equation x^2+2x+5=0. Use the method of completing the square.

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giorgiana1976 | College Teacher | Valedictorian

Posted June 18, 2010 at 2:55 AM (Answer #1)

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First, we'll add -5 bith sides:

x^2+2x = -5

The technique of completing the square is made in this way:

We'll divide the coefficient of the second term in x, by 2: 2/2=1.

Now, we'll raise to square this value: (1)^2 = 1

Now, we'll add 1, both sides:

x^2+2x+1 = -5+1

(x+1)^2 = -4

x+1 = +/-sqrt(-4)

x+1 = +/-i*sqrt4

x+1 = +/-2i

x1 = -1+2i

x2 = -1-2i

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neela | High School Teacher | Valedictorian

Posted June 18, 2010 at 3:29 AM (Answer #2)

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x^2+2x+5 = 0 .To solve by the method of completing the square.

Solution:

x^2+2x becomes complete square if we write x^2+2x+1.

So x^2+2x+5 = 0 could be written as:

(x^2+2x+1)+4 = 0. Or

(x+1)^2+4 = 0. Or

(x+1)^2 = -4. Or

(x+1) = +or- {sqrt(-4)}. Or

x1 = -1 +{sqrt(-4)} = -1+2i

x2 = -1-sqrt{-4} = -1-2i , where i = sqrt(-1).

 

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brianevery | High School Teacher | (Level 2) Adjunct Educator

Posted June 18, 2010 at 3:41 AM (Answer #3)

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Use complete the square to solve x^2+2x+5=0

To do this, I am going to use a method slightly different than is often taught. I am going to re-write the equation like this -

(x^2+2x+ ___)+5=0 now what number can I put in the blank that will make the expression in the parenthesis a perfect square? it has to be a number that is half of b (in ax^2+bx+c) and is the square of that number. b =2 and half of 2 is 1, 1 squared is 1 so the number I can substitute into the blank would be 1. Now I have added 1 to the left side of the equation, to keep it equal I have to either also subtract 1 from the left side, or add 1 to the right side of the equation. I choose to subtract it from the left side like this -

(x^2+2x+1)+5-1=0  Now what is in the parenthesis is a perfect square trinomial and it factors to (x+1)(x+1) or (x+1)^2 and my new equation simplified becomes

(x+1)^2+4=0 subtract 4 from both sides, and you get

(x+1)^2=(-4) take the square root of both sides to get

x+1=+/-sqrt(-4) re-write the right side to the product of sqrt(-1)*sqrt(4) which simplifies to (sqrt(-1) = i and sqrt(4)=2)=+/-2i this then gives you

x+1=+/-2i subtract 1 from both sides and the result is

x=+/-2i-1 which means that there are no real zero solutions to the equation, convention has us writing imaginary numbers in the form a+bi so we would re write our answer like this

x=-1+/-2i or

x=-1+2i, -1-2i

 

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