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Solving Quadratic Equations by Completing the Square

by Holly
(USA)











































solving quadratic equations by completing the square


HELP PLEASE


x^2+4+1=0

or like

x^2+x-1=0


thank you!!

Comments for Solving Quadratic Equations by Completing the Square

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Aug 13, 2011
Completing the Square
by: Staff

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Part II


(b) x² + x - 1 = 0


add 1 to each side of the equation to move the 1 to the right side of the equation

x² + x - 1 + 1 = 0 + 1

x² + x + 0 = 0 + 1

x² + x = 0 + 1

x² + x = 1


Divide the coefficient of x by 2

1/2 = 1/2

Square the result

(1/2)² = 1/4

Add this value to each side of the equation. This means that you will add ¼ to each side of the equation.

x² + x + ¼ = 1 + ¼

x² + x + ¼ = 1¼


THE REASON for moving the “1” to the right side of the equation, and then adding “¼” to each side of the equation is: CHANGE THE LEFT SIDE OF THE EQUATION SO IT CAN BE FACTORED AS A PERFECT SQUARE:

x² + x + ¼ = 1¼

(x + ½)(x + ½) = 1¼

(x + ½)² = 1¼


Take the square root of each side of the equation

√(x + ½)² = ±√(1¼)

x + ½ = ±√(1¼)


Subtract ½ from each side of the equation to remove the ½ from the left side of the equation. This leaves the variable x as the only term on the left side of the equation.

x + ½ - ½ = -½ ±√(1¼)

x + 0 = -½ ±√(1¼)

x = -½ ±√(1¼)

x = -½ ±√(5/4)

x = -½ ±[√(5)]/2


1st value of x₁ = -½ plus the square root of 1¼

x₁ = -½ + √(1¼)

x₁ ≈ -½ +

x₁ ≈ 0.6180339887499



2nd value of x₂ = -½ minus the square root of 1¼

x₂ = -½ - √(1¼)

x₂ ≈ -½ -

x₂ ≈ -1.6180339887499

x ∈{-1.6180339887499, 0.6180339887499}


the final answer to part ( b) is:

x = -½ ±√(1¼)

or

x = -½ ±√(5/4)

or

x = -½ ±[√(5)]/2

or

x ∈{-1.6180339887499, 0.6180339887499}


(1) Click the following link to VIEW the graphical solution; or (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/quad-eq-2011-08-13-b.png




check the solution by substituting the two numerical values of x into the original equation

for x₁ = 0.6180339887499


x² + x - 1 = 0


(0.6180339887499)² + (0.6180339887499) - 1 = 0

0.3819660112501 + 0.6180339887499 - 1 = 0

0 = 0, OK → x₁ = 0.6180339887499 is a valid solution



for x₂ = -1.6180339887499


x² + x - 1 = 0

(-1.6180339887499)² + (-1.6180339887499) - 1 = 0

2.6180339887499 - 1.6180339887499 - 1 = 0

0 = 0, OK → x₂ = -1.6180339887499 is a valid solution



Thanks for writing.

Staff
www.solving-math-problems.com


Aug 13, 2011
Completing the Square
by: Staff


The question:

Part I

by Holly
(USA)

solving quadratic equations by completing the square


HELP PLEASE


x^2+4+1=0

or like

x^2+x-1=0


thank you!!




The answer:


(a) x² + 4x + 1 = 0

subtract 1 from each side of the equation to move the 1 to the right side of the equation

x² + 4x + 1 – 1 = 0 - 1

x² + 4x + 0 = 0 – 1

x² + 4x = 0 – 1

x² + 4x = - 1


Divide the coefficient of 4x by 2

4/2 = 2

Square the result

(2)² = 4

Add this value to each side of the equation. This means that you will add 4 to each side of the equation.

x² + 4x + 4 = -1 + 4

x² + 4x + 4 = 3


THE REASON for moving the “1” to the right side of the equation, and then adding “4” to each side of the equation is: CHANGE THE LEFT SIDE OF THE EQUATION SO IT CAN BE FACTORED AS A PERFECT SQUARE:

x² + 4x + 4 = 3

(x + 2)(x + 2) = 3

(x + 2)² = 3


Take the square root of each side of the equation

√ (x + 2)² = ±√(3)

x + 2 = ±√(3)


Subtract 2 from each side of the equation to remove the 2 from the left side of the equation. This leaves the variable x as the only term on the left side of the equation.

x + 2 - 2 = -2 ±√(3)

x + 0 = -2 ±√(3)

x = -2 ±√(3)


1st value of x₁ = -2 plus the square root of 3

x₁ = -2 + √(3)

x₁ ≈ -2 + 1.7320508075689

x₁ ≈ -0.2679491924311


2nd value of x₂ = -2 minus the square root of 3

x₂ = -2 - √(3)

x₂ ≈ -2 - 1.7320508075689

x₂ ≈ -3.7320508075689


x ∈{-3.7320508075689, -0.2679491924311}


the final answer to part (a) is:

x = -2 ±√(3)

or

x ∈{-3.7320508075689, -0.2679491924311}


(1) Click the following link to VIEW the graphical solution; or (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/quad-eq-2011-08-13-a.png




check the solution by substituting the two numerical values of x into the original equation


for x₁ ≈ -0.2679491924311

x² + 4x + 1 = 0

(-0.2679491924311)² + 4(-0.2679491924311) + 1 = 0

0.0717967697245 - 1.0717967697244+ 1 = 0

0 = 0, OK → x₂ = -0.2679491924311 is a valid solution


for x₂ ≈ -3.7320508075689

x² + 4x + 1 = 0

(-3.7320508075689)² + 4(-3.7320508075689) + 1 = 0

13.928203230276 - 14.928203230276 + 1 = 0

0 = 0, OK → x₁ = -3.7320508075689 is a valid solution


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