121 - 4c² = 0
(11 + 2c)(11 - 2c) = 0
121-4c² is the difference of squares.
121-4c² = 11² - (2c)² = (11+2c) (11-2c)
recognize that these are both perfect squares, and the square root of 121 = 1, and the square root of 4c^2 = 2c.
FORMULA: DIFFERENCE OF SQUARES:
(a^2-b^2) = (sqrta^2 - sqrtb^2)(sqrta^2 + sqrtb^2) = (a-b)(a+b)
APPLICATION TO OUR EQUATION:
(121 - 4c^2) = (sqrt121 - sqrt4c^2)(sqrt121 + sqrt4c^2) = (11 - 2c)(11 + 2c)
121 -4c^2 = 0
(11)^2 -(2c)^2 = 0
a^2 -b^2 = 0
(a-b)(a+b)=0
a=11
b=2c
(11-2c)(11+2c) = 0
(11+2c)(11-2c) = 0
Use the difference of squares:
a^2 - b^2 = (a - b)(a + b)
so in this case:
121 - 4c^2 = 0
(11)^2 - (2c)^2 = 0
(11 - 2c)(11 + 2c) = 0 => factored form, if you need the roots then:
11 - 2c = 0
11 = 2c
c = 11/2
and
11 + 2c = 0
11 = -2c
c = -11/2
thus:
c = ± 11/2 are the roots of the equation.
I hope this helps.
(11 - 2c) ( 11 + 2c ) = 0
(11 + 2c)(11 - 2c)
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Answers & Comments
121 - 4c² = 0
(11 + 2c)(11 - 2c) = 0
121-4c² is the difference of squares.
121-4c² = 11² - (2c)² = (11+2c) (11-2c)
recognize that these are both perfect squares, and the square root of 121 = 1, and the square root of 4c^2 = 2c.
FORMULA: DIFFERENCE OF SQUARES:
(a^2-b^2) = (sqrta^2 - sqrtb^2)(sqrta^2 + sqrtb^2) = (a-b)(a+b)
APPLICATION TO OUR EQUATION:
(121 - 4c^2) = (sqrt121 - sqrt4c^2)(sqrt121 + sqrt4c^2) = (11 - 2c)(11 + 2c)
121 -4c^2 = 0
(11)^2 -(2c)^2 = 0
a^2 -b^2 = 0
(a-b)(a+b)=0
a=11
b=2c
(11-2c)(11+2c) = 0
(11+2c)(11-2c) = 0
Use the difference of squares:
a^2 - b^2 = (a - b)(a + b)
so in this case:
121 - 4c^2 = 0
(11)^2 - (2c)^2 = 0
(11 - 2c)(11 + 2c) = 0 => factored form, if you need the roots then:
11 - 2c = 0
11 = 2c
c = 11/2
and
11 + 2c = 0
11 = -2c
c = -11/2
thus:
c = ± 11/2 are the roots of the equation.
I hope this helps.
(11 - 2c) ( 11 + 2c ) = 0
(11 + 2c)(11 - 2c)