Solve: a) 2x2 – 7x + 2 = 0 by completing the square. b) 2x2 + 9x + 4 = 0 by factoring?

a)
2x² - 7x + 2 = 0
2x² - 7x = -2
2[x² - (7/2)x] = -2
x² - (7/2)x = -1
x² - 2(7/4)x + (7/4)² = -1 + (7/4)²
[x - (7/4)]² = -(16/16) + (49/16)
[x - (7/4)]² = 33/16
x - (7/4) = ±√(33/16)
x - (7/4) = ±(√33)/4
x = (7/4) ± (√33)/4
x = (7 + √33)/4 or x = (7 - √33)/4


b)
2x² + 9x + 4 = 0
(2x + 1)(x + 4) = 0
2x + 1 = 0 or x + 4 = 0
x = -1/2 or x = -4

a)
2x^2-7x+2 = 0
2(x^2 - (7/2) x ) + 2 = 0
2( x^2 - 2(7/4) x + (7/4)^2 - (7/4)^2 ) + 2 = 0
2( (x^2 - 2(7/4) x + (7/4)^2 ) - (7/4)^2 ) + 2 = 0
2( (x-7/4)^2 - (7/4)^2 ) + 2 = 0
2( (x-7/4)^2 - 49/16) + 2 = 0
2(x-7/4)^2 - 49/8 + 2 = 0
2(x-7/4)^2 -49/8+16/8 = 0
2(x-7/4)^2 -33/8 =0
2(x-7/4)^2 = 33/8
(x-7/4)^2 = 33/16
x-7/4 = +/- sqrt(33) / 4
x = 7/4 + sqrt(3) / 4
x = 7/4 - sqrt(33) /4
x= (7 + sqrt(33)) /4
x = (7- sqrt(33))/4

b)
2x^2+9x+4 = 0
2x^2+8x+x+4=0
2x(x+4)+1(x+4)=0
(x+4)(2x+1) = 0
x+ 4 = 0
x = -4
2x+1 = 0
2x=-1
x= -1/2

x= -1/2 or -4

a)
2 [ x² - 7x/2 ] = - 2
2 [ x² - 7x/2 + 49/4 ] = - 2 + 49/2
2 [ x - 7/2 ]² = 45/2
[ x - 7/2 ]² = 45/4
x - 7/2 = ± √45 / 2
x = 7/2 ± √45 / 2
x = [ 7 ± √45 ] / 2
x = [ 7 ± 3√5 ] / 2

b)
[ 2x + 1 ] [ x + 4 ] = 0
x = - 1/2 , x = - 4

a) 2x2 – 7x + 2 = 0 (I prefer yo keep away from fractions early on; so try this trick:)
16x^2 - 2 * 7 * 4x = -16
16x^2 - 2 * 7 * 4x + 49 = -16 + 49 = 33
So (4x - 7)^2 = 33,
and 4x = 7 ± √33
So x = (7 ± √33) / 4

b) 2x2 + 9x + 4 = 0
2 * 4 + 1 * 1 = 9
so (2x + 1) (x + 4) = 0
So x = -1/2 or -4