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Old 09-16-2014   #1
Dragonballful23

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Default Quadratic Easy

Easy quadratics (solve for x)


x^2-3x+2

x^2-4x+2

2x^2-5x+2
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Old 09-22-2014   #2
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[QUOTE=Dragonballful23;44096]Easy quadratics (solve for x)


1) x^2-3x+2

Splitting middle term of quadratic

x^2-x-2x+2

x(x-1)-2(x-1)

(x-1)(x-2)

x=1 and x=2.



2)x^2-4x+2

The Quadratic Formula:

For ax^2 + bx + c = 0, the value of x is given by:

x = [ -b sqrt(b^2 - 4ac) ] / 2a

a=1 , b= -4 , c= 2

x = [ 4 sqrt(-4^2 - 4*1*2) ] / 2*1

x = [ 4 sqrt(16 - 8) ] / 2

x = [ 4 sqrt(8) ] / 2

x = [ 4 2 sqrt(2) ] / 2

x = [ 2 sqrt(2) ]

x = [ 2 + sqrt(2) ] and x = [ 2 - sqrt(2) ]



3) 2x^2-5x+2

Splitting middle term of quadratic

2x^2-4x-x+2

2x(x-2)-1(x-2)

(x-2)(2x-1)

x = 2 and x = 1/2

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Old 09-27-2014   #3
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Quote:
Originally Posted by Dragonballful23 View Post
Easy quadratics (solve for x)


x^2-3x+2

x^2-4x+2

2x^2-5x+2
x^2-3x+2 = (x-2)(x-1) = 0, or the solutions are x = 2,1
x^2-4x+2 can't be factored, so by the quadratic formula it's
x = 4 +/- 8 all over 2, or x = 2 +/- root 2

2x^2 -5x+2 can be factored as (2x-1)(x-2) = 0, or x = 1/2, 2
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Old 09-29-2014   #4
Dragonballful23

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Quote:
Originally Posted by Pi= View Post
x^2-3x+2 = (x-2)(x-1) = 0, or the solutions are x = 2,1
x^2-4x+2 can't be factored, so by the quadratic formula it's
x = 4 +/- 8 all over 2, or x = 2 +/- root 2

2x^2 -5x+2 can be factored as (2x-1)(x-2) = 0, or x = 1/2, 2
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