Quadratic equation using the formula help needed?

ACTUALLY, we usually call quadratic to any function where the variable has a power of 2, it is the formula to express the function of a parabola. The quadratic formula is used to solve most commonly for the variable x (the parabola then opens up or down), but there are also parabolas which open right and left.
To solve for x means that you are finding the values where the parabola intercepts the x axis.

The general equation for the parabola is:

y= ax^2 + bx + c

Sometimes you can solve this by factoring. When that is too hard you can use the complete the square method, and finally the easiest method: the quadratic formula, where the 2 values of x can be found by inserting the coefficients imbued in the equation given to you. Usually we get 2 values because the x is raised to the 2nd power.

x1,2= [ -b +- sqr( b^2- 4ac)]/2a

All you need to do is plug in the equation the values given:

Example :

y= 3x² + 10x – 2 ............(Your problem above)

When you make it equal to 0, what you are saying is that you need to find the values of x that make the function have a zero value. the intersections with the x-axis.

So a= 3
b= 10

and c= -2

Now you plug in those numbers in the quadratic formula:

x1,2= [ -(10) +- sqr( 10^2 - (4)(3)(-2) ] /(2(3)

x1,2= [-10+ - (12)] / 6

or x1= 11/3= 1/3=0.33
and x2= -11/3 =-3.66

Now lets go and see if what I wrote above is true: Go to the image-graph and notice the intersections in the x axis coincide:

http://mathway.com/answer.aspx?p=grap?p=ySMB013xSMB07(2)+10x-2?p=True?p=True?p=True?p=False?p=True?p=-4?p=2?p=-11?p=3

Now you do the next one.

ANSWER

a)
3x^2 + 10x - 2 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 3
b = 10
c = -2

x = [-10 ±√(100 + 24)]/6
x = [-10 ±√124]/6
x = [-10 ±√(2^2 * 31)]/6
x = [-10 ±2√31]/6
x = [-5 ±√31]/3

∴ x = [-5 ±√31]/3

= = = = = = = =

b)
4x^2 - 4x - 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 4
b = -4
c = -1

x = [4 ±√(16 + 16)]/8
x = [4 ±√32]/8
x = [4 ±√(4^2 * 2]/8
x = [4 ±4√2]/8
x = [1 ±√2]/2

∴ x = [1 ±√2]/2

Quadratic formula:
When ax^2 + bx + c = 0, than roots are
x1 = (-b + sq.root(b^2 - 4ac)) / 2a
and
x2 = (-b - sq. root(b^2 - 4ac)) / 2a
So in this problem we have:
1.) 3x^2 + 10x -2 folows that a =3, b = 10 nad c = -2.
So roots of this equations are:
x1 = (-10 + sq.root(100 - 4*3*(-2))) / 2*3
x1 = (-10 + sq.root(100+24)) / 6
x1 = (-10 + sq.root(4*31)) / 6
x1 = (-10 + sq.root4 *sq.root (31) ) / 6
x1 = (-10 + 2*sq.root(31)) /6
x1 = -10/6 + (2/6)*sq.root(31)
x1 = -4/3 + (1/3)*sq.root(31)
In the same way
x2 = (-10 - sq.root(100 - 4*3*(-2))) / 2*3, indeed
.
.
.
x2 = (-4/3) - (1/3)*sq.root(31)
2.) 4x^2 - 4x -1
x1 = (-(-4) + sq.root(4^2 - 4*4*(-1))) / 2*4
x1 = (4 + sq.root(32)) / 8
x1 = (4 + sqroot(16*2)) / 8
x1 = (4 + sq.root(16)*sq.root(2)) / 8
x1 = (4 + 4*sq.root(2)) / 8
x1 = 4/8 + (4/8)*sq.root(2)
x1 = 1/2 + (1/2)*sq.root(2)
Analogiclly:
x2 = 1/2 - (1/2)*sq.root(2)

a) 3x² + 10x – 2 = 0

a=3 b= 10 c= -2

Quadratic Formula

x = [- b ± √ (b² - 4ac)] / 2a
=> x = [- 10 ± √ (10² - 4(3)(-2)] / 2(3)
=> x = [- 10 ± √ (124] / 6
use a calculator
etc

Part a)
x = [ - 10 ± √ (100 + 24) ] / 6
x = [ - 10 ± √ (124) ] / 6
x = [ - 10 ± 2√ (31) ] / 6
x = [ - 5 ± √ (31) ] / 3

Part b)
x = [ 4 ± √ (16 + 16) ] / 8
x = [ 4 ± √ (32) ] / 8
x = [ 4 ± 4√ (2) ] / 8
x = [ 1 ± √ (2) ] / 2