Solving Quadratics?

Question

Could someone please help me with these! Thank You very much to all that reply!

4x2 - 16 = 0

2x2 = 16

x2 + 5 = 0

x2 + 5x - 6 = 0

x2 - 25 = 0

3x2 = -9x

2x2 - 4x + 1 = 0

3x2 + 2x = 7

x2 + 8x + 9 = -7

Answer 1

1)
4x² - 16 = 0
(2x+4)(2x-4) = 0
(x+2)(x-2) = 0

x= -2 ou x = 2

2)
2x² = 16
x² = 8
x= 8^¹/2 ou.. x=-8^¹/2

3)
x² + 5 = 0
x² = -5
x = (5^¹/2)i ..ou.. x = (-5^¹/2)i

4)
x² + 5x - 6 = 0

x = [-5±(5²+4*1*6)^¹/2] / 2
x = [-5±(25+24)^¹/2] / 2
x = [-5±7] / 2

x = -6 ..ou.. x = 1

5)
x² - 25 = 0
x² = 25
x = -5 .. ou .. x = 5

6)
3x² = -9x
x² = -3x
x = -3 .. ou .. x = 0

7)
2x² - 4x + 1 = 0

x = [4±((-4)²-4*2*1)^¹/2] / (2*2)
x = [4±(16-8)^¹/2] / 4
x = [4±8^¹/2] / 4
x = [4±(2*2^¹/2)] / 4
x = [2±2^¹/2] / 2
x = [2-2^¹/2] / 2 .. ou .. x = [2+2^¹/2] / 2

8)
3x² + 2x = 7
3x² + 2x - 7 = 0

x = [2±(2²+4*3*7)^¹/2] / (2*3)
x = [2±(4+84)^¹/2] / 6
x = [2±2*22^¹/2] / 6
x = [1±22^¹/2] / 3
x = [1-22^¹/2] / 3 .. ou .. x = [1+22^¹/2] / 3

9)
x² + 8x + 9 = -7
x² + 8x + 16 = 0
(x+4)² = 0
x = -4

₢

Answer 2

Let's see...

4x2 -16 = 0 => (2x + 4)(2x - 4) = 0 => x = 2 or x =-2

2x2 = 16 => x =2

x2 +5 = 0 it has no solution in the real numbers.

x2 + 5x -6 = 0 => (x -1)(x+6) => x = 1 or x = -6

x2 -25 = 0 => (x+5)(x-5) = 0 => x = 5 or x = -5

3x2 = -9x => 3x2 +9x = 0 => 3x(x+3) = 0 => x = 0 or x=-3

2x2 -4x +1 = 0 => x = (-b +- (sqrt(b2-4ac))/2a

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

gives as result:

(4 +-(sqrt(16-4.2.1)))/2.2 = (4 +- sqrt(8) ) /4 = 1 +- sqrt(2)/2

x = 1 + sqrt(2)/2 (aprox 1.7) or x = 1-sqrt(2)/2 (aprox 0.3)

3x2 +2x -7 = 0 just as the previous...

(-2 +- sqrt(4-4.3.(-7)))/2.3 =
(-2 +- sqrt(88))/6 =
(-2 +- 2.sqrt(22))/6 =

x = -1.89 or x = 1.23

x2 +8x +9 = -7 => x2 +8x+16 = 0 => (x+4)(x+4) =0 => x = -4

Answer 3

4x^2-16 = 0, 4x^2 = 16, x^2 = 4, thus x = 2

2x^2 = 16, x^2 = 16/2, x^2 = 8, thus x = 4

x^2+5 = 0, x^2 = -5, thus x = sqrt( -5 )

x^2+5x-6 = 0, By factoring: (x-1)(x+6) = 0, thus x = 1 and x = -6

x^2-25 = 0, x^2 = 25 thus x = 5

This all I have time to solve but it will get you started...

Answer 4

1) 4x^2-16=0 4x^2=16 x^2=4 x=+ or -2
2) 2x^2=16 x^2=8 x=root 8
3) x^2+5=0 x^2= -5 no real solutions
4) x^2+5x-6=0 (x-1)(x+6)=0 x= 1 or -6
5)x^2-25=0 x^2=25 x=+ or - 5
6) 3x^2=-9x 3x^2+9x=0 x^2+3x=0 using -b+or- root b^2-4xaxc/2a x=-0.38 or -2.6
7) 2x^2-4x+1=0 using same formula x= 1.71 or 3.41
8) 3x^2+2x=7 3x^2+2x-7=0 formula again & x= 1.85 or -2.52
9) x^2+8x+9=-7 x^2+8x+2=0 same way x=0.26 or -7.74

Ok, 10points

Answer 5

4x^2 - 16 = 0

This is a difference of squares, so you factor it as follows:
(2x - 4)(2x + 4) = 0
2x = 4 or 2x = -4
x = 2 or x = -2

2x^2 = 16
x^2 = 8
When taking the square root of both sides, you have to put a "plus or minus" (or +/-), so

x = +/- sqrt(8)

You do the rest.

Answer 6

4x^2-16=0
4x^2=16
x^2=4
x= +or - 2

2x^2=16
x^2=8
x= + or - 2sqr(2)

x^2+5=0
x^2= -5
x= no sulution

x^2+5x-6
(x-1)(x+6)
x= 1 and -6

x^2-25-0
x^2=25
x= 5 or -5

3x^2= -9x
3x^2 +9x
3x(x+3)
x= 0 or -3

Answer 7

these equations can all be solved using the quadratic formula

for all ax^2 + bx + c = 0

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

Answer 8

x=2
x=2sqrt(2)
x=sqrt(-5)
x==-6,x=1
x=5
x=-3
x=4+/-2sqrt(2)/2
x=-2+/-2sqrt(5)i/3
x=-8+/-2sqrt(14)/2