Solving quadratics: 4x^2 = 1, 11x^2 - 12 + 1 = 0, etc.

rsalo1981

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Jan 12, 2008
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1
I am sooo confused with these problems. Could I please get some help? Thank you!

1. 4x^2=1

2. 11x^2-12+1=0

3. x^2-12=0
 

TchrWill

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Jul 7, 2005
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DIvide through by 4 yielding x^2 = 1/4
Take the square root of both sides yielding x = 1/2

11x^2-12+1=0
My apologies for assuming an x after the 12 in my earlier response.
SUBHOTOSH KAHN corrected my error.
11x^2 - 11 = 0
11x^2 = 11
x^2 = 1
x = +1 or -1

Adding 12 to both sides yields x^2 = 12
Taking the square root of both sides yields x = sqrt12 = sqrt(4(3)) = 2sqrt3 = 3.4641
 

Subhotosh Khan

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Jun 18, 2007
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Re: HELPPPPP please

TchrWill said:
4x^2=1
DIvide through by 4 yielding x^2 = 1/4
Take the square root of both sides yielding x = 1/2

11x^2-12+1=0
Using the quadratic equation,
x = [12+/-sqrt(12^2 - 4(11))]/22
x = [12+/-sqrt(100)]/22
x = [12+10]/22 = 1 or [12-10]/22 = 1/11
Above is not correct - as posted:

\(\displaystyle 11x^2 -12 + 1 = 0\)

\(\displaystyle 11x^2 - 11 = 0\)

\(\displaystyle 11x^2 = 11\)

\(\displaystyle x^2 = 1\)

\(\displaystyle x = \pm 1\)
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x^2-12=0
Adding 12 to both sides yields x^2 = 12
Taking the square root of both sides yields x = sqrt12 = sqrt(4(3)) = 2sqrt3 = 3.4641[/quote]
 
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