two quick quadratic formula question?

Question

Use the quadratic formula to slove these two equations

1) 6x²-x-12=0

2) x²-6x+13=0

Answer 1

The quadratic formula is: [-b+/-((b^2-4ac))/(2a)] where a is the number befor x^2, b is the number before x and c is the integer. Using this formula we get:

1. x = [1+/- (1-(4)(6)(-12))^1/2/((2)(6))]
x = [1+/- 17]/(12)
x = 18/12 or -16/12
x = 1.5, x = -4/3

2. x = [6+/- (36-(4)(1)(13))^1/2/((2)(1))]
x = [6 +/- (16)^(1/2) i}(2) {since there is a negative sign inside the sq root we replace it by and positive sq root and imaginary number}.
x = [6 +/- 4i]/(2)
x = 3 + 2i, x = 3-2i

Answer 2

the quadratic quation is of the main formula: ax^2+bx+c=0
1) 6x^2-x-12=0 then a=6, b=-1 , c=-12
we have D=b^2- 4ac=1-(-288)=289 (D is delta) and D>0 then we obtain 2 distinct roots in R(real #s)
then x=(-b-radicalD)/2a=(1-17)/12=-16/12=-4/3=-1.3
or x=(-b+radicalD)/2a=(1+17)/12=18/12=3/2=1.5
then x=-1.3 or x=1.5

2) a=1, b=-6 , c=13
D=36-52=-16<0
then we obtain no roots in R

Answer 3

1) 6x²-x-12=0
6x² -9x + 8x -12 = 0
3x(2x - 3) + 4 (2x -3) = 0
x = 3/2, -4/3

2) x²-6x+13=0
(6 * 6) - 4(13) (1) = 36 -52 = -16 < 0

(This equation cannot be solved)

Answer 4

1) solve this by using the method of SPLITTING THE MIDDLE TERM.....
6x²-x-12=0
6x² - 9x + 8x - 12 = 0
3x ( 2x - 3 ) + 4 (2x - 3 ) =0
( 3x +4 ) ( 2x - 3 ) =0

so x can be either -4/3 or 3/2..


2)
the equation x²-6x+13=0 is in the form of equation ax²+ bx + c=0

& the formula for solving such equation is

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


apply this formula to this eq. n u'l get the answer as

3+ 2i & 3 - 2i.....where i means that the roots are complex roots..............


mail me if u still have probs

Answer 5

1. ax²+bx+c=0......here, a=6, b=(-1) and c=(-12)
(+- is plus or minus and sqrt is square root)

x= (-b +- sqrt(b²-4ac)) / 2a)

x= (1+- sqrt((-1)² - (4)(6)(-12))) / (2)(6)

x= (1+- sqrt(1 + 288)) / 12

x= (1 +- 17) / 12
_
x= 18/12= 1.5 or x= -16/12 = -1.3

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2. ax²+bx+c=0......here, a=1, b=(-6) and c=(13)
(+- is plus or minus and sqrt is square root)

x= (-b +- sqrt(b²-4ac)) / 2a)

x= (-1 +- sqrt((-6)² - (4)(1)(13))) / (2)(1)

x= (-1 +- sqrt(36-52)) / 2

x= (-1 +- sqrt(-16)) / 2........remember that -16 factors into 16 * -1...the sqrt of 16 is 4 and the sqrt of -1 is "i" the imaginary number)

x= (-1 +- 4i) / 2

x= (-1 + 4i) / 2= -0.5 + 2i or x= (-1 - 4i) / 2= -0.5 + 2i

Answer 6

6x^2 - x - 12 = 0

x = (-(-1) ± sqrt((-1)^2 - 4(6)(-12)))/(2(6))
x = (1 ± sqrt(1 + 288))/12
x = (1 ± sqrt(289))/12
x = (1 ± 17)/12
x = (18/12) or (-16/12)
x = (3/2) or (-4/3)

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x^2 - 6x + 13 = 0

x = (-(-6) ± sqrt((-6)^2 - 4(1)(13)))/(2(1))
x = (6 ± sqrt(36 - 52))/2
x = (6 ± sqrt(-16))/2
x = (6 ± 4i)/2
x = 3 ± 2i
x = 3 - 2i or 3 + 2i

Answer 7

Quadratic Formula:

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

1) x = 1.5 or -1.33
This can also be solved by factoring.
(2x - 3)(3x +4) = 0

2) no solution

Answer 8

1) [1 +/- sr(329)]/12 sr=square root of ( )
2) there is no possible real number answer.
(you would have to square root a negative number)