Algebra? Word Problem?

Question

Decide all of the values of b in the following equations that will give one or more real number solutions.
(a) 3x^2+bx-3=0
(b)5x^2+bx+1=0
(c)-3x^2+bx-3=0

Answer 1

using the quadratic equation:
x=-b+/-Sqrt(b^2-4ac)/2a

a) a=3 c=-3
plug in:
x=-b+/-sqrt(b^2-4*3*(-3))/6
6x=-b+/-sqrt(b^2+36)
the question is real solutions, so
b^2+36>=0
b^2>=-36
all real values of b

b)
b^2-4ac>=0
b^2>=4(5)(1)
b^2>=20
b can be less than or equal to -sqrt(20)
or b can be greater than or equal to sqrt(20)

c)
b^2>=4(-3)(-3)
b^2>=36
-6<=b b>=6

j

Answer 2

a) divide the equation by 3, factor to (x -1)(x+1) = 0, because the middle term does not exist, bx/3 = 0, so b = 0 or x = 0 (anything multiplied by 0 is 0). X cannot = 0 because the original equation would be solved to 0 = -3, which is a no-no.

b) as with A, factor the equation to (5x + 1)(x + 1) = 0
OR (5x-1)(x-1)=0, solve for x, replace the values to get b

c) first, divide the equation by -3, then factor the equation, sovle for X then replace values to get B.

Answer 3

If the students current grade is a 90% then we’re of course assuming it’s out of 100% 90/100 Now since the students grade improved 20% from the last grading period then you add that to the previous 100% giving you 90/ 120 = .75 And since grades are in percentages you multiply .75 by 100 giving you 75% as the students previous grade

Answer 4

1, -1

Answer 5

You want to get b by itself so first you have to solve for x. Then once you have x solved, you can replace it in the equation and find what b will be. It should be coordinates I believe.

Answer 6

use the quadratic equation and find the range of b for the root to be zero or postive

Answer 7

a. all real numbers
b. all real numbers
c. all real numbers

since y0u will square s0mething in each number.. y0u will always have a p0sitive number. all 0f the number y0u will substitute t0 b will still make the answer real!=)

Answer 8

a- b=0
b- b=6
c- b= -10 or -9

i think...
you will have to check for yourself