Quadratic equations ???

Question

(1) x^2 = 5x + 2

(2) 4x^2 -3x + 3 = 0

(3) 3x^2 = 11x + 4


HELP !!! :)

Answer 1

First get your equation into proper form: ax² + bx + c = 0.

There are different ways to solve quadratics. I'll use a different method for each one:

1) Try this with completing the square
x² - 5x - 2 = 0
x² - 5x = 2
x² - 5x + 25/4 = 2 + 25/4
(x - 5/2)² = 33/4
x - 5/2 = ±(√33)/2
x = (5±√33)/2
x = (5+√33)/2 and x = (5-√33)/2

2) Try this with the quadratic forumula,
x = (-b±√(b²-4ac)) / 2a

Plug in the values and you get
(3±√(9-4*4*3)) / 8. But this gives you a negative value under the square root, meaning that the equation has no real solutions.

3) You can factor out 3x² - 11x - 4 by inspection, or "reverse F.O.I.L.".
The only factors of 3 are 1 and 3, so the first terms of the factors you get look like:
(3x ---- )(x ----)
The "4" at the end in negative, so you need to have one "+" and one "-" inside. The middle term is negative, so the minus should be in the second term so that it gets multiplied by 3:
(3x + ---- )(x - ----)
Finally, the factors of 4 are (1,4) and (2,2). The only set-up that will get us -11x is:
(3x + 1)(x - 4)
So setting this equal to zero, x = -1/3 or 4.

Answer 2

The general equation is ax^2 + bx + c = 0

1) x^2 = 5x + 2
x^2 - 5x - 2 = 0

a = 1, b=-5, c=-2

2) 4x^2 - 3x + 3 = 0
already in the proper form

a=4, b=-3, c=3

3) 3x^2 = 11x + 4
3x^2 - 11x - 4

a=3, b=-11, c=-4

Now plug in values of a, b and c into the quadratic equation

Answer 3

1. First: set the equation to "0" - subtract 5x from both sides...

x^2 - 5x = 5x - 5x + 2
x^2 - 5x = 2

*Subtract 2 from both sides...

x^2 - 5x - 2 = 2 - 2
x^2 - 5x - 2 = 0

Sec: there are three coefficients for three variables...
a = 1 > b = -5 & > c = -2

*Replace the values with the corresponding variables in the Quadtratic Equation...

x = [- b +/- V`b^2 - 4ac)] / 2a

x = [- (-5) +/- V`(-5)^2 - 4(1)(-2)] / 2(1)

x = [- (-5) +/- V`(-5)^2 - 4(1)(-2)] / 2

x = [5 +/- V`(-5)^2 - 4(1)(-2)] / 2

x = [5 +/- V`(-5)(-5) - 4(1)(-2)] / 2

x = [5 +/- V`25 - 4(-2)] / 2

x = [5 +/- V`25 + 8] / 2

x = [5 +/- V`33] / 2

Fourth: there are two solutions, one is positive-the other is negative...

a. x = [5 + V`33] / 2
x = 5/2 + (V`33)/2

b. x = [5 - V`33] / 2
x = 5/2 - (V`33)/2

*Now-try the rest :-)

Answer 4

x^2 = 5x + 2

Quadratic Formula: [ -b +or- square root of (b^2-4ac) ] / 2a.

Equation A) x^2 -5x -2 = 0
a = 1, b = -5, c = -2

[ -(-5) +or- square root of (-5)^2 -4(1)(-2) ] / 2(1)


x1 = ( 5 + square root of 33 ) / 2

x2 = ( 5 - square root of 33 ) / 2

Check answers by substituting x1 and x2 into
....Equation A): x^2 - 5x - 2 = 0

The rest of your problems can be solved the same way. =)

Answer 5

Set them Equal to Zero first.

1) -x^2 +5x+2 =zero

2) is already set equal to zero

3) -m3^2+11x+4=0

Now follow the fomula x=( b +or- the Square root of b^2-4ac)/2a

With Ax+By+C=0