For people good in math easy ten points?

Question

Find the distance between (-2,6) and (1,3)

Find the value of the discriminant of the equation 25x^2 - 30x +9 = 0

Answer 1

what is the discriminant in the quadratic formula.....its the part under the radical so it's b^2-4(a)(c)

in this equation a=25 b=30 and c=9

so just plug the correct values in

30^2-4(25)(9)

30^2= 900

-4(25)= -100

-100(9)= -900

so 900-900=0

what does this tell us.....that the equation has a single real solution  

Answer 2

Distance formula = Sqr root of (x2 - x1)^2 + (y2 - y1)^2

Given points (-2,6) and (1,3)

D = Sqr root of (1 - (-2))^2 + (3 - 6)^2 = (3)^2 + (-3)^2 =9 +9 = Sqr root of 18 = Sqr root of 9 x 2 =

3 radical 2

The discriminant is part of the quadratic equation.

The discriminant (D) = b^2 - 4ac

1)If D > 0, the equation has 2 real solutions.

2)If D = 0, the equation has 1 real solution.

3)If D < 0, the equation has 2 conjugate imaginary solutions.

25x^2 - 30x +9 = 0

a=25, b= -30, c=9

b^2 - 4ac = (-30)^2 - 4(25)(9) = 900 - 900 = 0

Condition #2 is met, so there is just one real solution.

x = - b / 2a

x = 30/2(25) = 30/50 = 3/5  

Answer 3

You don't have to be good in Math. You just need to know the formulas.

for pts (a,b) and (c,d), distance = sqrt[(a-c)^2 + (b-d)^2]

discriminant of ax^2 + bx + c = 0 is b^2 - 4ac.

Answer 4

using the distance formula,

d = √(6-3)²+(-2-1)² = √18 = 3√2

discriminant = b²-4ac = 900-25(9) = 900 - 225 = 675

that's it! ;)

Answer 5

For the first one, use the distance formula:

d = sqrt( (x_2 - x_1)^2 + (y_2 - y_1)^2 )

For the second one, find the discriminant using:

Delta = b^2 - 4ac

Your polynomial is: 25x^2 - 30x + 9 = ax^2 + bx + c