If the distance between point r(a, a,a) and point j(6,-2,0) is 10, then the value of a could be?

Step-by-step explanation:

Let's use the distance formula here.

The distance between R(a, a, a) and J(6, -2, 0) is 10.  The whole formula is:

d = √(6 - a)^2 + (-2 - a)^2 + (0 - a)^2 = 10.

We must solve for a.

Note that d² = 100.

Thus, we have d² = 100 = (6 - a)² + (-2 - a)² + (-a)²

Expanding the squares:

                              100 = 36 - 12a + a² +  4 + 2a + a² + a²

Combining the constants:

                               100 - 36 - 4 = 60

Combining the a terms:  -12a + 2a = -10a;

Combining the a² terms:  3a²

Then the sum 100 = 36 - 12a + a² +  4 + 2a + a² + a² becomes

                          60 = -10a + 3a²

Rewrite this in standard form for a quadratic:

                           3a² - 10a - 60 = 0

Here the coefficients are a = 3, b = -10 and c = -60, so the discriminant, b²-4ac is 100-4(3)(-60), or 100 + 720, or 820.  

                -(-10) ± √820)

Then a =

                        2(3)

or:

                  10 ± √4·√205

Then a =

                             6

                   5  ± √205

Then a =

                           3  

greater than one

step-by-step explanation: