Use the quadratic formula to find the exact solution.?

Question

x^2 - 8x + 65 = 0

Answer 1

the ans my frend is x= 13 & x= -5

Answer 2

(x - 13) (x + 5) = 0

either x = 13, or x = -5

8 was a typo.. corrected now.

Answer 3

The quadratic solution is:

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

So, subbing your numbers,

x = [-(-8) +- sqrt((-8)^2 - 4(1)(65))] / 2(1)

x = [8 +- sqrt(64 - 260)] / 2

x = [8 +- sqrt(-196)]/2

Now, depending on whether you are using complex numbers or not (you did not specify), you may have to stop.  Because the determinant (the part inside the square root) is negative, there on no real roots for the given equation (parabola).

However, with complex numbers, two answers can be resolved.  Let's continue, using the substitution sqrt(-1) = i

x = [8 +- sqrt(-1)sqrt(196)]/2

x = [8 +- i(14)]/2

x = 4 +- 7i

x = 4 + 7i or x = 4 - 7i

There is the exact answer.  Again, this only works if you are using complex numbers.  If you are using real numbers, like most of the math in high school, there is not real answer.

Answer 4

a=1

b=-8

c=65

x=  { 8   ± Sq Root ( 64  -  260 ) }  ÃƒÂƒÃ‚·  2

Answer 5

a=1

b=-8

c=65

x=(-b+-sqrt(b²-4ac))/2a

x=(8+-sqrt((-8)²-4(1)(65)))/2(1)

x=(8+-sqrt(64-260))/2

x=(8+-sqrt(-196))/2

x=(8+-sqrt(49*4)i)/2

x=(8+-14i)/2

x=4+-7i

x=4+7i  and x=4-7i

I hope this can be useful

David

Answer 6

This equation is of form ax^2+bx+c

a = 1 b = -8 c = 65

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

x=[8 +/-sqrt(-8^2-4(1)(65)]/(2)(1)

discriminant is b^2-4ac =-196

i^2 = -1, so √i^2 = i

No real roots: The complex roots are

x=[8 +i √(196)] / (2)(1)

x=[8 -i √(196)] / (2)(1)

x=[8+i14] / 2

x=[8-i14] / 2

4+7i and 4-7i

Answer 7

4-i7, 4+i7

Answer 8

To solve this quadratic equation for quadratic or general formula, the formula in order that it realizes is this one:

....-b±√b²-4ac

x=----------------

...........2a

The values of "a", "b" and "c", according to the quadratic form ax² +bx +c, they are these:

x² - 8x + 65 = 0

a = 1

b= -8

c = 65

We replace all these values in the formula, this way:

....-(-8)±√(-8)²-4(1)(65)

x=----------------------------

..................2(1)

....8±√64-260

x=----------------

...........2

....8±√-196

x=------------

.........2

To extract root squared to-196, we have to extract first with the property of radicación to √-1 that it is equal to imaginary number, symbol "i", that hereinafter will be replaced, this way:

....8±√-1√196

x=-----------------

............2

....8±i√196

x=-------------

...........2

Now if we can extract the square root of 196 (positive number) this way:

....8±i√196

x=--------------

............2

....8±14i

x=---------

........2

Now we look for both possible values, then we simplify and obtain the result, this way:

.......8+14i....2(4+7i)

x1=---------=------------=4+7i

...........2.............2

.......8-14i....2(4-7i)

x1=---------=------------=4 - 7i

...........2.............2

We extract the common factor to reduce easier and to come to the results. I hope that my explanations use you as help, bye-bye =).