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Solve the polynomial equation using special products (x-3)^2=36.

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singwriter | Student, Grade 11 | eNoter

Posted July 28, 2011 at 1:19 AM via web

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Solve the polynomial equation using special products (x-3)^2=36.

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giorgiana1976 | College Teacher | Valedictorian

Posted July 28, 2011 at 1:23 AM (Answer #1)

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We notice that if we'll subtract both sides 36, we'll get a difference of two squares:

(x-3)^2 - 36 = 0

We know that the difference of two squares returns the product:

a^2 - b^2 = (a-b)(a+b)

Let a = x - 3 and b = 6

(x-3)^2 - 36 = (x - 3 - 6)(x - 3 + 6)

We'll cancel each factor:

(x - 3 - 6) = 0

x - 9 = 0

x = 9

(x - 3 + 6) = 0

x + 3 = 0

x = -3

The solutions of the polynomial equation are {-3 ; 9}.

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samhouston | Middle School Teacher | (Level 1) Associate Educator

Posted July 28, 2011 at 1:25 AM (Answer #2)

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(x - 3)^2 = 36

This is an example of the Perfect Squares Pattern, a type of special product.  The binomial (x - 3)^2 = 36, therefore (x - 3) must equal either 6 or -6 because both numbers squared will equal 36.  So you solve x - 3 = 6 and x - 3 = -6.

x - 3 = 6

x = 9

 

x - 3 = -6

x = -3

 

Solutions:  {-3, 9}

 

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shaznl1 | High School Teacher | Salutatorian

Posted July 28, 2011 at 3:22 AM (Answer #3)

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The restriction of this equation is to use special products.

From the equation

(x-3)^2=36

We can see that both sides are prefect squares.

Thinking back to stone age(not literally), we remember the difference of squares formula:

a^2-b^2= (a+b)(a-b)

Subtract 36 from both sides

(x-3)^2-36=0

using the formula

(x-3-6)(x-3+6)=0

(x-9)(x+3)=0

x+3=0 or x-9=0

x=-3 or 9

Solutions

X=-3, 9

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Jyotsana | Student, Grade 10 | Valedictorian

Posted February 15, 2014 at 1:56 AM (Answer #4)

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(x-3)^2

Both Method

x*x=x^2

x*-2=-3x

-3*x=-3x

-3*-3=9

x^2-3x-3x+9=36

x^2-6x+9=36

x^2-6x+9-36=0

x^2-6-27=0

Quadratic Formula

(-6+Square Root(-6)^2-4(1)(-27))/4(1)

x=-3

0r x=9

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